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Ramanujan J (2014) 35:21–110
DOI 10.1007/s11139-013-9528-5
Rediscovery of Malmsten’s integrals, their evaluation
by contour integration methods and some related
results
Iaroslav V. Blagouchine
Received: 18 November 2012 / Accepted: 5 October 2013 / Published online: 7 January 2014
© Springer Science+Business Media New York 2014
Abstract This article is devoted to a family of logarithmic integrals recently treated
in mathematical literature, as well as to some closely related results. First, it is shown
that the problem is much older than usually reported. In particular, the so-called
Vard i ’ s i n t egral, which is a particular case of the considered family of integrals, was
first evaluated by Carl Malmsten and colleagues in 1842. Then, it is shown that un-
der some conditions, the contour integration method may be successfully used for
the evaluation of these integrals (they are called Malmsten’s integrals). Unlike most
modern methods, the proposed one does not require “heavy” special functions and is
based solely on the Euler’s Γ-function. A straightforward extension to an arctangent
family of integrals is treated as well. Some integrals containing polygamma functions
are also evaluated by a slight modification of the proposed method. Malmsten’s inte-
grals usually depend on several parameters including discrete ones. It is shown that
Malmsten’s integrals of a discrete real parameter may be represented by a kind of
finite Fourier series whose coefficients are given in terms of the Γ-function and its
logarithmic derivatives. By studying such orthogonal expansions, several interesting
theorems concerning the values of the Γ-function at rational arguments are proven.
In contrast, Malmsten’s integrals of a continuous complex parameter are found to be
connected with the generalized Stieltjes constants. This connection reveals to be use-
ful for the determination of the first generalized Stieltjes constant at seven rational
arguments in the range (0,1)by means of elementary functions, the Euler’s con-
stant γ, the first Stieltjes constant γ1and the Γ-function. However, it is not known
if any first generalized Stieltjes constant at rational argument may be expressed in
the same way. Useful in this regard, the multiplication theorem, the recurrence rela-
tionship and the reflection formula for the Stieltjes constants are provided as well.
A part of the manuscript is devoted to certain logarithmic and trigonometric series
I.V. Blagouchine (B)
University of Toulon, La Valette du Var (Toulon), France
e-mail: iaroslav.blagouchine@univ-tln.fr
22 I.V. Blagouchine
related to Malmsten’s integrals. It is shown that comparatively simple logarithmico–
trigonometric series may be evaluated either via the Γ-function and its logarithmic
derivatives, or via the derivatives of the Hurwitz ζ-function, or via the antiderivative
of the first generalized Stieltjes constant. In passing, it is found that the authorship of
the Fourier series expansion for the logarithm of the Γ-function is attributed to Ernst
Kummer erroneously: Malmsten and colleagues derived this expansion already in
1842, while Kummer obtained it only in 1847. Interestingly, a similar Fourier series
with the cosine instead of the sine leads to the second-order derivatives of the Hur-
witz ζ-function and to the antiderivatives of the first generalized Stieltjes constant.
Finally, several errors and misprints related to logarithmic and arctangent integrals
were found in the famous Gradshteyn & Ryzhik’s table of integrals as well as in the
Prudnikov et al. tables.
Keywords Logarithmic integrals ·Logarithmic series ·Theory of functions of a
complex variable ·Contour integration ·Rediscoveries ·Malmsten ·Var d i ·Number
theory ·Gamma function ·Zeta function ·Rational arguments ·Special constants ·
Generalized Euler’s constants ·Stieltjes constants ·Otrhogonal expansions
Mathematics Subject Classification 33B15 ·30-02 ·30D10 ·30D30 ·11-02 ·
11M35 ·11M06 ·01A55 ·97I30 ·97I80
1 Introduction
1.1 Introductory remarks and history of the problem
In an article which appeared in the American Mathematical Monthly at the end of
1980s, Vardi [67] treats several interesting logarithmic integrals found in Gradshteyn
and Ryzhik’s tables [28]. His exposition begins with the integrals
π/2
ˆ
π/4
lnln tg xdx =
1
ˆ
0
lnln 1
x
1+x2dx =∞
ˆ
1
lnln x
1+x2dx =1
2
∞
ˆ
0
lnx
chxdx
=π
2lnΓ(3/4)
Γ(1/4)√2π,(1)
which can be deduced one from another by a simple change of variable (these formu-
las are given in no. 4.229-7, 4.325-4 and 4.371-1 of [28]). Although the first results
for such a kind of integrals may be found in the mathematical literature of the 19th
century (e.g., in famous tables [62]), they continue to attract the attention of modern
researchers and their evaluation still remains interesting and challenging. Vardi’s pa-
per [67] generated a new wave of interest to such logarithmic integrals and numerous
works on the subject, including very recent ones, appeared since [67], [2], [13], [7],
[47], [46], [68], [4], [8], [44]. On the other hand, since the subject is very old, it is
hard to avoid rediscoveries. For the computation of the above mentioned integral (1),
modern authors [2], [13, p. 237], [7, p. 160], [46], [4], [44], send the reader to the
Malmsten’s integrals and their evaluation by contour integration 23
Fig. 1 A fragment of p. 12 from the Malmsten et al.’s dissertation [40]
Vardi’s paper [67].1Vardi, failing to identify the author of formula (1) and failing
to locate its proof, proposed a method of proof based essentially on the use of the
Dirichlet L-function. However, formula (1), in all four forms, was already known to
David Bierens de Haan [62, Table 308-28, 148-1, 260-1], and if we go to a deeper
exploration of this question, we find that integral (1) was first evaluated by Carl Jo-
han Malmsten2and colleagues in 1842 in a dissertation written in Latin [40,p.12],
see Fig. 1. A part of this dissertation was later republished in the famous Journal für
die reine und angewandte Mathematik6[41], see, e.g., p. 7 for integral (1). Moreover,
two other logarithmic integrals,
1
ˆ
0
lnln 1
x
1+x+x2dx =∞
ˆ
1
lnln x
1+x+x2dx =π
√3lnΓ(2/3)
Γ(1/3)
3
√2π(2a)
and
1
ˆ
0
lnln 1
x
1−x+x2dx =∞
ˆ
1
ln ln x
1−x+x2dx =2π
√3ln6
√32π5
Γ(1/6)(2b)
mentioned in [2,67], [13, p. 238], [4,70], were also first evaluated by Malmsten
et al. [40, pp. 12 and 43] and [41, formulas (12) and (72)].3Malmsten and his col-
leagues evaluated many other beautiful logarithmic integrals4and series as well, but
1Only Bassett [8], an undergraduate student, remarked that solutions for integrals (1)and(2a,2b)aremuch
older than Vardi’s paper [67].
2Carl Johan Malmsten, written also Karl Johan Malmsten (born April 9, 1814 in Uddetorp, died Febru-
ary 11, 1886 in Uppsala), was a Swedish mathematician and politician. He became Docent in 1840, and
then, Professor of mathematics at the Uppsala University in 1842. He was elected a member of the Royal
Swedish Academy of Sciences (Kungliga Vetenskaps–akademien) in 1844. He was also a minister without
portfolio in 1859–1866 and Governor of Skaraborg County in 1866–1879. For further information, see
[36, vol. 17, pp. 657–658].
3Both results are presented here in the original form, as they appear in the given sources. In fact, both
formulas may be further simplified and written in terms of Γ(1/3)only, see (45)and(44), respectively
(see also exercise no. 32 where both integrals appear in a more general form). Surprisingly, the latter fact
escaped the attention of Malmsten, of his colleagues and of many other researchers. Moreover, Vardi [67,
p. 313] even wrote that “in (2a) the number 3 plays the ‘key role’ and in (2b) 6 is the ‘magic number”’.
A more detailed criticism of the latter statement is given in exercise no. 30.
4Many of which were independently evaluated in [2]andin[44].
24 I.V. Blagouchine
unfortunately, none of the above-mentioned contemporary authors mentioned them.
Moreover, the latter even named integral (1) after Vardi (they call it Vardi’s i n t egral),
and so did many well-known internet resources such as Wolfram MathWorld site [70]
or OEIS Foundation site [57].
On the other hand, it is understandable that, sometimes, it can be quite difficult
to find the original source of a formula, especially because of the oldness of the
result, because the chain of references may be too long and confusing, and because it
could be published in many different languages. For example, Gradshteyn and Ryzhik
wrote originally in Russian (their book [28] is a translation from Russian), Bierens de
Haan published usually in Dutch or in French, Malmsten wrote in Swedish, French
and Latin, and we now use mostly English. As the reference for (1), Gradshteyn and
Ryzhik [28] as well as Vardi [67], cite the famous Bierens de Haan’s tables [62]. In
the latter, on the p. 207, Table 148, the reference for the integral (1) is given as “(IV,
265)”.5This means that this result comes from the 4th volume of the Memoirs of the
Royal Academy of Sciences of Amsterdam, which is entirely composed of the Tables
d’intégrales définies by Bierens de Haan [61], and which is an old version of the well-
known Nouvelles tables d’intégrales définies [62]. The old version [61] is essentially
the same as the new one [62], except that it provides original sources (the new version
[62] contains much less misprints and errors, but original references given in the
old edition were removed). Thus, we may find in the old edition [61, pp. 264–265,
Table 191-1] that integral (1) was evaluated by Malmsten in the work referenced as
“Cr. 38. 1.” This is often the most difficult part of the work, to understand what an old
abbreviation may stand for. Bierens de Haan does not explain it, and unfortunately,
neither do modern dictionaries nor encyclopedia. After several hours of search, we
finally found that “Cr.” stands for “Crelle’s Journal”, which is a jargon name for
the Journal für die reine und angewandte Mathematik.6The number 38 stands for
the volume’s number, and 1 is not the issue’s number but the number of the page
from which the manuscript starts. Furthermore, a deeper study of Malmsten’s works,
see e.g. [63, p. 31], shows that this article is a concise and updated version of the
collective dissertation [40] which was presented at the Uppsala University in April–
June 1842.7Therefore, taking into account the undoubted Malmsten and colleagues’
priority in the evaluation of the logarithmic integrals of the type (1) and (2a), (2b),
we think that integral (1) should be called Malmsten’s integral rather than Vard i ’s
integral. Throughout the manuscript, integrals of kind (1) and (2a), (2b) are called
Malmsten’s integrals.
The aim of the present work is multifold; accordingly, the article is divided in three
parts. In the first part (Sect. 2), we present Malmsten’s original proof that Vardi and
other modern researchers missed. The presentation of this proof may be of interest
5Another frequently encountered notation for references in Bierens de Haan’s tables [61]and[62]—which
may be not easy to understand for English-speaking readers—is “V. T.” which stands for voir tableau,
i.e. “see table” in English.
6English translation: “Journal for Pure and Applied Mathematics”.
7However, we were surprised to see that Malmsten in the article [41] did not even mention the afore-
mentioned dissertation [40]. In fact, integrals (1)and(2a), (2b) were very probably derived by one of his
students or colleagues, but now it is almost impossible to know who exactly did it.
Malmsten’s integrals and their evaluation by contour integration 25
for a large audience of readers for multiple reasons. First, it may be quite difficult
to find references [40] and [41], as well as cited works. Second, the manuscript was
written in Latin, and references are in French. Latin, being discarded from the study
program of most mathematical faculties, may be difficult to understand for many
researchers. Third, Malmsten does not make use of special functions other than Γ-
function; instead, he smartly employs elementary transformations, so that his proof
may be understood even by a first-year student. Fourth, the work [41] contains numer-
ous misprints in formulas and a proper presentation might be quite useful as well. At
the end of the presentation, we briefly discuss further Malmsten et al.’s contributions,
such as, for example, the Fourier series expansion for the logarithm of the Γ-function
(obtained 5 years before Kummer) or the derivation of the reflection formula for two
series closely related to ζ-functions (obtained 17 years before the famous Riemann’s
functional relationship for the ζ-function). Also, connections between the logarithm
of the Γ-function and the digamma function, written by Malmsten as a kind of dis-
crete cosine transform, are interesting and provide some further ideas that we later
re-used in Sect. 4.5. At the end of this part, we remark that several widely known
tables of integrals, such as Gradshteyn and Ryzhik’s tables [28], Bierens de Haan’s
tables [61,62], and probably, Prudnikov et al.’s tables [53], borrowed a large amount
of Malmsten’s results, butin most of them, original references to Malmsten were lost.
Moreover, some of these results appear with misprints.
In the second part of the manuscript (Sect. 3), we introduce a family of logarith-
mic integrals of which integral (1) and many other Malmsten’s integrals are simple
particular cases. We propose an alternative method for the analytical evaluation of
such a kind of integral. Unlike most modern methods, the proposed one does not
require “heavy” special functions and is based on the methods of contour integra-
tion. A non-exhaustive condition under which considered family of integrals may be
always expressed in terms of the Γ-function is provided. A straightforward exten-
sion to an arctangent family of integrals is treated as well. At the end of this part,
we consider in detail examples of application of the proposed method to four most
frequently encountered Malmsten’s integrals.
The third part of this work (Sect. 4) is designed as a collection of original exercises
containing new formulas and theorems, which can be derived directly or indirectly by
the proposed method. The exercises and theorems have been grouped thematically:
•Logarithmic Malmsten’s integrals containing hyperbolic functions and some
closely related results are treated in Sect. 4.1. In particular, integrals, which can be
evaluated by the direct application of the proposed method, are given in Sect. 4.1.1.
These may be roughly divided in two parts: relatively simple Malmsten’s integrals
containing two or three parameters (e.g. exercises no. 1,2,4,5,6-a, 7,8,17,
etc.) and complete Malmsten’s integrals depending on three or more parameters,
including discrete ones (e.g. exercises no. 3,6-b,c,d, 9,13,11,14). Simple Malm-
sten’s integrals, some of which are evaluated up to order 20,18 often lead to various
special constants such as Euler’s constant γ,Γ(1/3),Γ(1/4),Γ(1/π), Catalan’s
constant G, Apéry’s constant ζ(3)and others. As regards complete Malmsten’s
integrals, whose evaluation is carried out up to order 4, it is found that such inte-
grals, when depending on a discrete real parameter, may be represented by a kind
of finite Fourier series whose coefficients are given in terms of the Γ-function and
26 I.V. Blagouchine
its logarithmic derivatives. In contrast, when the considered discrete real parameter
becomes continuous and complex, such integrals may be expressed by means of
the first generalized Stieltjes constants (such exercises are placed in Sect. 4.5).
•The results closely related to logarithmic Malmsten’s integral are placed in
Sect. 4.1.2. These include the evaluation of integrals similar to Malmsten’s ones
and that of certain closely connected series. Most of these series are logarithmico–
trigonometric and may be evaluated either via the Γ-function and its logarithmic
derivatives, or via the derivatives of the Hurwitz ζ-function, or via the antideriva-
tive of the first generalized Stieltjes constant (conversely, such series may be re-
garded as Fourier series expansions of the above-mentioned functions).
•In Sect. 4.2, we treat ln ln-integrals, most of which are obtained by a simple change
of variable of integrals from Sect. 4.1. In the same section, we also show that
Vardi’s hypothesis about the relationship between the argument of the Γ-function
and the degree in which the poles of the corresponding integrand are the roots of
unity is not true in general (exercise no. 30).
•Section 4.3 is devoted to arctangent integrals. Similarly to logarithmic integrals,
arctangent integrals can be roughly classified into several categories: compara-
tively simple, complete and closely related (such as, e.g., exercise no. 40 where
an analog of the second Binet’s formula for the logarithm of the Γ-function is
derived).
•In Sect. 4.4, we show that some slight modifications of the method developed in
Sect. 3may be quite fruitful for the evaluation of certain integrals containing log-
arithm of the Γ-function and the polygamma functions.
•Lastly, in Sect. 4.5 we put exercises and theorems related to the values of the Γ-
function at rational arguments and to the Stieltjes constants. Mostly, these results
are deduced from precedent exercises. For instance, by means of finite orthogonal
representations obtained for the Malmsten’ integrals in Sect. 4.1, we prove several
interesting theorems concerning the logarithm of the Γ-function at rational argu-
ments, including some variants of Parseval’s theorem (exercises no. 58–62). By the
way, with the help of the same technique one can derive similar theorems implying
polygamma functions. In the second part of Sect. 4.5, we show that some com-
plete Malmsten’s integrals, which were previously evaluated in Sect. 4.1, may be
also expressed by means of the first generalized Stieltjes constants. This connec-
tion between Malmsten’s integrals of a real discrete and of a continuous complex
parameters is not only interesting in itself, but also permits evaluation of the first
generalized Stieltjes constant γ1(p) at p=1
2,1
3,1
4,1
6,2
3,3
4,5
6by means of el-
ementary functions, the Euler’s constant γ, the first Stieltjes constant γ1and the
Γ-function (see exercise no. 64). However, it is still unknown if any first gener-
alized Stieltjes constant at rational argument may be expressed in the same way
(from this point of view, the evaluation of γ1(1/5)could be of special interest). In
this framework, we also discovered that the sum of the first generalized Stieltjes
constant γ1(p),p∈(0,1), with its reflected version γ1(1−p) may be expressed,
at least for seven different rational values of p, in terms of elementary functions,
the Euler’s constant γand the first Stieltjes constant γ1. At the same time, it is
not known if other sums γ1(p) +γ1(1−p) share the same property. An alternative
evaluation of integrals from exercises no. 65–66 could probably provide some light
on this problem.
Malmsten’s integrals and their evaluation by contour integration 27
Finally, answers for all exercises were carefully verified numerically with Maple
12 (except exercises with Stieltjes constants which were verified with Wolfram Al-
pha Pro). By default, if nothing is explicitly said, the presented result coincides with
the numerical one. Actually, only in few cases Maple 12 fails to correctly evaluate
integrals. For instance, it fails both numerically and symbolically to evaluate the first
integral on the left in (1). Maple 12.0 gives (−πln2 +iπ2)/4≈0.544 +i2.467,
while it is clear that this integral has no imaginary part at all, and the real one is nei-
ther correctly evaluated. By the way, authors of [44] also reported incorrect numerical
and symbolical evaluation of this integral by Mathematica 6.0. However, unlike [44],
we will not specify wherever Maple 12 is able or unable to evaluate integrals analyt-
ically, because in almost all cases Maple 12 was unable to do it.
1.2 Notations
Throughout the manuscript, following abbreviated notations are used: γ=
0.5772156649... for the Euler’s constant, γnfor the nth Stieltjes constant, γn(p) for
the nth generalized Stieltjes constant at point p,8G=0.9159655941 ... for Cata-
lan’s constant, xfor the integer part of x,tgzfor the tangent of z,ctgzfor the cotan-
gent of z,chzfor the hyperbolic cosine of z,shzfor the hyperbolic sine of z,thzfor
the hyperbolic tangent of z,cthzfor the hyperbolic cotangent of z.9In order to avoid
any confusion between compositional inverse and multiplicative inverse, inverse
trigonometric and hyperbolic functions are denoted as arccos, arcsin, arctg,... and
not as cos−1,sin
−1,tg
−1,.... We write Γ(z),Ψ(z),Ψ
1(z), Ψ2(z), Ψ3(z), . . . , Ψn(z)
to denote, respectively, gamma, digamma, trigamma, tetragamma, pentagamma,...,
(n −2)th polygamma functions of argument z. The Riemann ζ-function, the η-
function (known also as the alternating Riemann ζ-function) and the Hurwitz ζ-
function are, respectively, defined as
ζ(s)=∞
n=1
1
ns,η(s)=∞
n=1
(−1)n+1
ns,ζ(s,v)=∞
n=0
1
(n +v)s,
v=0,−1,−2,..., with Re s>1fortheζ-functions and Res>0fortheη-function.
Where necessary, these definitions may be extended to other domains by the principle
of analytic continuation. For example, one of the most known analytic continuations
for the Hurwitz ζ-function is the so-called Hermite representation
ζ(s,v)=v1−s
s−1+v−s
2+2∞
ˆ
0
sinsarctg x
v
(e2πx −1)v2+x2s/2dx , Re v>0,(3)
which extends ζ(s,v) to the entire complex plane except at s=1, see e.g. [9,vol.I,
p. 26, Eq. 1.10(7)]. Note also that the η-function may be easily reduced to the Rie-
8We remark, in passing, that by convention γn≡γn(1)for any natural n.
9Most of these notations come from Latin, e.g “ch” stands for cosinus hyperbolicus, “sh” stands for sinus
hyperbolicus,etc.
28 I.V. Blagouchine
mann ζ-function η(s) =(1−21−s)ζ(s), while the Hurwitz ζ-function is an indepen-
dent transcendent (except some particular values). Moreover, the alternating Hurwitz
ζ-function η(s,v) may be similarly reduced to the ordinary Hurwitz ζ-function
η(s,v) =∞
n=0
(−1)n
(n +v)s=lim
z→s21−zζz, v
2−ζ(z,v),(4)
v= 0,−1,−2,...,Res>0. Rezand Im zdenote, respectively, real and imaginary
parts of z. Natural numbers are defined in a traditional way as a set of positive inte-
gers, which is denoted by N. Kronecker symbol of arguments land kis denoted by
δl,k. Letter iis never used as index and is √−1. Complex integration over region A
Im zBmeans that the complex line integral is taken around an infinitely long hor-
izontal strip delimited by inequality AIm zB, where (A, B ) ∈R2(i.e. the inte-
gration contour is a rectangle with vertices at [R+iA,R+iB,−R+iB,−R+iA]
with R→∞. The notation resz=af(z) stands for the residue of the function f(z) at
the point z=a. Other notations are standard. Finally, we remark that the references
to the formulas are given between parentheses “( )”, those to the number of exercise
from Sect. 4are preceded by “no.”; the bibliographic references are given in square
brackets “[ ]”.
2 Malmsten’s method and its results
2.1 Malmsten’s original proof of the integral formula (1)
The proof is presented in a way closest to the original [41]; we have only replaced the
old notations by the new ones, as well as correcting numerous misprints in formulas
(by the way, [40] contains much less misprints). In the effort to make it more accessi-
ble for the readers, several modern references were also added, but, of course, the old
ones are also left. These references are marked with an ∗(only in this subsection).
Malmsten begins with the elementary integral
∞
ˆ
0
e−xz sin vz dz =v
x2+v2,x>0,v>0,
which, being integrated10 over vfrom 0 to u, becomes
2∞
ˆ
0
(1−cosuz) e−xz
zdz =lnx2+u2−2lnx. (5)
10This operation is permitted because the considered improper integral is uniformly convergent with re-
spect to v[34, p. 44, § 1.12]*, [58, vol. II, pp. 262–269]*, [10, pp. 175–179]*.
Malmsten’s integrals and their evaluation by contour integration 29
The latter logarithm is then replaced by one of Frullani’s integrals [34, pp. 406–407,
§ 12.16]∗,[50]∗,[24, p. 455]∗,
lnx=∞
ˆ
0
e−z−e−xz
zdz, (6)
which yields
lnx2+u2=2∞
ˆ
0
e−z−e−xz cosuz
zdz. (7)
Multiply the last equality by sh au
sh πu, parameter abeing in the range (−π, +π), and
then integrate it over all values of ufrom 0 to ∞
∞
ˆ
0
sh au
shπu lnx2+u2du =2∞
ˆ
0∞
ˆ
0
sh au
sh πue−z−e−xz cos uzdudz
z.(8)
In virtue of formulas (b) and (a) from [64, vol. II, p. 186]11
∞
ˆ
0
sh au
sh πu du =1
2tg a
2and ∞
ˆ
0
sh au
sh πu cosuz d u =sin a
2(chz+cos a) ,
where −π<a<πand −∞ <z<∞, expression (8) may be rewritten as follows:
∞
ˆ
0
sh au
sh πu lnx2+u2du =∞
ˆ
0tg a
2−2e−xz sin a
1+2e−zcosa+e−2ze−zdz
z,(9)
−π<a<π. Now make a change of variable in the last integral by putting y=e−z.
This yields
∞
ˆ
0
sh au
sh πu lnx2+u2du =
1
ˆ
0tg a
2−2yxsin a
1+2ycosa+y2dy
ln 1
y≡Ta(x), (10)
−π<a<π, where the last integral was designated by Ta(x) for brevity. It can be
easily demonstrated that if the parameter ais chosen so that a=πm/n, numbers m
and nbeing positive integers such that m<n(in other words, if ais a rational part of
π), then, the integral on the right part of the last equation may be always expressed
11These formulas may be also derived by contour integration methods, see e.g. [59, pp. 186, 197–198]∗,
[69, p. 132]∗,[23, pp. 276–277]∗.
30 I.V. Blagouchine
in terms of the Γ-function. The differentiation of Ta(x) with respect to xgives
dT
dx =2
1
ˆ
0
yxsin a
1+2ycosa+y2dy. (11)
But if the parameter ais a rational part of π, the latter integral, in virtue of what was
established in [64, vol. II, pp. 163–165], is
1
ˆ
0
yxsin a
1+2ycosa+y2dy
=⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
n−1
l=1
(−1)l−1sin(la ) Ψx+n+l
2n−Ψx+l
2n,if m+nis odd,
1
2(n−1)
l=1
(−1)l−1sin(la ) Ψx+n−l
n−Ψx+l
n,if m+nis even.
By substituting these formulas into the right part of (11), and by calculating the an-
tiderivative, we obtain
Ta(x) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
C1+2
n−1
l=1
(−1)l−1sin(la ) lnΓx+n+l
2n
Γx+l
2n,if m+nis odd,
C2+2
1
2(n−1)
l=1
(−1)l−1sin(la ) lnΓx+n−l
n
Γx+l
n,if m+nis even,
(12)
where C1and C2are constants of integration. In order to find them, the following
procedure is adopted. Put in the last formula, first x=r, and then x=s. Subtracting
one from another and dividing by minus two, we have a(r, s) ≡1
2[Ta(s) −Ta(r )]=
1
ˆ
0
yr(1−ys−r)sin a
1+2ycosa+y2·dy
ln 1
y
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
n−1
l=1
(−1)l−1sin(la ) lnΓs+n+l
2nΓr+l
2n
Γr+n+l
2nΓs+l
2n,if m+nis odd,
1
2(n−1)
l=1
(−1)l−1sin(la ) lnΓs+n−l
nΓr+l
n
Γr+n−l
ns+l
n,if m+nis even.
Malmsten’s integrals and their evaluation by contour integration 31
The difference between a(0,1)and a(1,2)yields
1
ˆ
0
(1−2y+y2)sin a
1+2ycosa+y2·dy
ln 1
y
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
n−1
l=1
(−1)l−1sin(la ) ln⎧
⎨
⎩
Γ2n+l+1
2nΓl+2
2nΓl
2n
Γ2l+1
2nΓn+l
2nΓn+l+2
2n⎫
⎬
⎭,
if m+nis odd,
1
2(n−1)
l=1
(−1)l−1sin(la ) ln⎧
⎨
⎩
Γ2n−l+1
nΓl+2
nΓl
n
Γ2l+1
nΓn−l
nΓn−l+2
n⎫
⎬
⎭,
if m+nis even.
Over and over again from (12)forx=1, we get
1
ˆ
0
(1−2y+y2)sin a
1+2ycosa+y2·dy
ln 1
y=(1+cosa)
×
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
C1+2
n−1
l=1
(−1)l−1sin(la ) ln⎧
⎨
⎩
Γ1+n+l
2n
Γ1+l
2n⎫
⎬
⎭,if m+nis odd,
C2+2
1
2(n−1)
l=1
(−1)l−1sin(la ) ln⎧
⎨
⎩
Γ1+n−l
n
Γ1+l
n⎫
⎬
⎭,if m+nis even.
By comparing last two expressions, one may easily identify both constants of inte-
gration:
⎧
⎪
⎪
⎨
⎪
⎪
⎩
C1=sin a·ln2n
1+cosa=tg a
2·ln2n,
C2=sin a·lnn
1+cosa=tg a
2·lnn.
In the final analysis, the substitution of these values into (12) yields
Ta(x) =∞
ˆ
0
sh au
sh πu lnx2+u2du
32 I.V. Blagouchine
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
tg πm
2n·ln2n+2
n−1
l=1
(−1)l−1sin πml
n·ln⎧
⎨
⎩
Γ1
2+x+l
2n
Γx+l
2n⎫
⎬
⎭,
if m+nis odd,
tg πm
2n·lnn+2
1
2(n−1)
l=1
(−1)l−1sin πml
n·lnΓ1−l−x
n
Γl+x
n,
if m+nis even,
(13)
where a≡πm/n. Now, one can easily deduce formula (1).12 Putting m=1 and
n=2in(13), we obtain
∞
ˆ
0
ln(u2+x2)
2ch
1
2πudu =2ln⎧
⎨
⎩
2Γx+3
4
Γx+1
4⎫
⎬
⎭.
By making a suitable change of variable in the above integral, and by taking into
account that the definite integral of ch−1xover x∈[0,∞)equals π/2, the latter
equation takes the form
∞
ˆ
0
ln(u2+x2)
chudu =2πln ⎧
⎨
⎩
√2πΓ
x
2π+3
4
Γx
2π+1
4⎫
⎬
⎭.(14)
Setting x=0 yields immediately formula (1) in its hyperbolic form.
By using a similar procedure and with the help of previous results, Malmsten also
evaluated
La(x) ≡∞
ˆ
0
chau
chπu lnx2+u2du (15)
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
sec πm
2n·ln2n+2
n
l=1
(−1)l−1cos (2l−1)mπ
2n·ln⎧
⎨
⎩
Γ1
2+2x+2l−1
4n
Γ2x+2l−1
4n⎫
⎬
⎭,
if m+nis odd,
sec πm
2n·lnn+2
1
2(n−1)
l=1
(−1)l−1cos (2l−1)mπ
2n·ln⎧
⎨
⎩
Γ1−2l−1−2x
2n
Γ2l−1+2x
2n⎫
⎬
⎭,
if m+nis even,
12From here, we shorten Malmsten’s proof since the result is almost straightforward. Malmsten’s proof
was actually longer because he aimed for more general formulas.
Malmsten’s integrals and their evaluation by contour integration 33
where a≡πm/n. The reader can find this result on the p. 41 [40, formula (46)] and
onthep.28[41, formula (70)].
In some particular cases, right parts of (13) and (15) may be largely simplified,
and sometimes, they even can be expressed in terms of only elementary functions.
For example,
Tπ
2(1)=∞
ˆ
0
ln(1+u2)
ch1
2πudu =4ln 2Γ(1)
Γ(1/2)=2ln 4
π,(16)
or
T2π
33
2=∞
ˆ
0
sh2
3πu
shπu ln9
4+u2du =√3⎧
⎨
⎩ln 6 +ln Γ11
12 Γ13
12
Γ5
12 Γ7
12 ⎫
⎬
⎭
=√3ln1
2ctg π
12,
or more general integrals
Tmπ
nn
2=∞
ˆ
0
shm
nπu
sh πu lnn2
4+u2du
=
n−1
l=1
(−1)l−1sin πml
n·lnn
2−lctgπ
4−πl
2n,
and
Lmπ
nn
2=∞
ˆ
0
ch(m
nπu)
chπu lnn2
4+u2du
=
n
l=1
(−1)l−1cos (2l−1)mπ
2n·lnn+1
2−lctgπ
4−π(2l−1)
4n,
which hold both only for m+nodd and where logarithms in right parts should be
regarded as limits when the index l=n/2 and l=(n +1)/2 respectively, see [40,
formulas (13), (21), (26), (55)],13 [41, formulas (8), (18), (23), (79)], [62, Table 258-
1,6,10,9], [61, Table 275-1,10,16,15]. Many other logarithmic integrals—which can
be expressed in a closed form—may be also found in [40] and [41].
At the end of this historical excursion, it may be of interest to remark that Legendre
was not far from Malmsten’s formula (1) in its hyperbolic form. On p. 190 [64,vol.II]
13There is a misprint in formula (13): the sign “−” in the denominator of the integrand should be replaced
by “+”.
34 I.V. Blagouchine
we find
∞
ˆ
0
dx
(eπx +e−πx)(m2+x2)=1
4mΨm
2+3
4−Ψm
2+1
4,
where mis not necessarily integer. Multiplying both sides by 2mand computing the
antiderivative yields
∞
ˆ
0
ln(m2+x2)
eπx +e−πx dx =lnΓm
2+3
4−lnΓm
2+1
4+C.
Notwithstanding, the constant of integration Cis not easy to determine, and this task
was achieved, albeit differently, by Malmsten and colleagues (C=1
2ln2), see also
comments on pp. 25–26 in [40]. In like manner, the second Binet’s formula for the
logarithm of the Γ-function may be also derived from Legendre’s work [64] (see, for
more details, exercise no. 40).
2.2 Brief discussion of other results obtained by Malmsten and his colleagues
Many other similar results were obtained by Malmsten and colleagues. Of course,
they noticed that the hyperbolic form of the integrand could be transformed into that
containing ln ln xor ln ln 1
xin the numerator, and many valuable results for such inte-
grals were obtained as well. Among these results, the most magnificent and remark-
able are perhaps these two:
1
ˆ
0
xn−2lnln 1
x
1+x2+x4+···+x2n−2dx =∞
ˆ
1
xn−2lnln x
1+x2+x4+···+x2n−2dx (17)
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
π
2ntg π
2nln2π+π
n
n−1
l=1
(−1)l−1sin πl
n·ln⎧
⎨
⎩
Γ1
2+l
2n
Γl
2n⎫
⎬
⎭,n=2,4,6,...
π
2ntg π
2nlnπ+π
n
1
2(n−1)
l=1
(−1)l−1sin πl
n·lnΓ1−l
n
Γl
n,n=3,5,7,...
see [40,p.12],[41,p.7]or[61, Table 191-5], [62, Table 148-4], [28, no. 4.325-9],
and
1
ˆ
0
xn−2lnln 1
x
1−x2+x4−···+x2n−2dx =∞
ˆ
1
xn−2lnln x
1−x2+x4−···+x2n−2dx (18)
=π
2nsec π
2n·lnπ+π
n·
1
2(n−1)
l=1
(−1)l−1cos (2l−1)π
2n·ln⎧
⎨
⎩
Γ1−2l−1
2n
Γ2l−1
2n⎫
⎬
⎭
Malmsten’s integrals and their evaluation by contour integration 35
holding for n=3,5,7,...,see[40, p. 42, Eq. (48)] (the latter result does not appear in
other sources). Particular cases of these formulas were rediscovered several times in
different forms by various authors, see e.g. no. 3.1–3.2, 3.5–3.614 [2] or examples7.3,
7.5 [44]. Moreover, Malmsten’s integrals (1) and (2a), (2b) are themselves particular
cases of the above integrals.
Malmsten and colleagues also treated some integrals having continuous powers of
the logarithm in the numerator and denominator. Evidently, only in few cases could
authors evaluate such integrals in a closed-form. However, such integrals permitted,
inter alia, to obtain the ln ln-integrals as a derivative with respect to the power of the
logarithm:
1
ˆ
0
P(x)ln ln 1
x
Q(x) dx =lim
a→0d
da
1
ˆ
0
P(x)lna1
x
Q(x) dx,
where P(x)and Q(x ) denote polynomials in x.
An important part of both Malmsten’s works [40] and [41] is also devoted to cer-
tain logarithmic series, to series related to ζ-functions and to some infinite products.
Among the results concerning logarithmic series, the most striking is, with no doubts,
the evaluation of the series of the kind
∞
n=1
sin an ·lnn
n,0<a<2π, (19)
see Fig. 2(for more details, see also exercise no. 20). This result, known as the Fourier
series expansion for the logarithm of the Γ-function, is usually (and erroneously)
attributed to Ernst Kummer who derived this expansion only in 1847, i.e. 5 years
later. Another important result in the field of series concerns certain infinite sums
related to ζ-functions. The famous reflection formula for the ζ-function
ζ(1−s) =2ζ(s)Γ (s)(2π)−scos πs
2,s= 0.(20)
is well-known and is usually attributed to Riemann who derived it in 1859 [54], [19,
p. 861], [31,p.23],[71, p. 269]. At the same time, it is much less known that Malm-
sten and colleagues derived analogous relationships for two other “similar” series
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
M(s) ≡2
√3
∞
n=1
(−1)n+1
nssin πn
3,M(1−s) =2
√3M(s)Γ (s)3s(2π)−ssin πs
2,
L(s) ≡∞
n=0
(−1)n
(2n+1)s,L(1−s) =L(s)Γ (s)2sπ−ssin πs
2,
(21)
14It seems, however, that there is a misprint in exercise no. 3.6 [2]. In the first line, the term x5should be
removed from the denominator of the integrand.
36 I.V. Blagouchine
Fig. 2 Fragments of pp. 62 (top) and 74 (bottom) from the Malmsten et al.’s dissertation [40], Cdesig-
nating the Euler’s constant γ. Writing a−πinstead of ain the former series yields (19), which is the
principal term in the Fourier series expansion of the logarithm of the Γ-function
Fig. 3 Bottom of p. 23 from the Malmsten et al.’s dissertation [40]. To p, we see the reflection formula for
M(s);bottom,forL(s). Moreover, similar reflection formulas were later derived by Malmsten in [41]for
other series similar to the η-function
0<s<1, already in 1842, see Fig. 3. The function L(s ) is directly related to the
alternating Hurwitz ζ-function L(s) =2−sη(s, 1
2), and therefore, Malmsten’s func-
tional equation may be rewritten as
η1−s, 1
2=2ηs, 1
2Γ(s)(2π)−ssin πs
2,
which holds even when s→0, if the right part is regarded as a limit. The similitude to
Malmsten’s integrals and their evaluation by contour integration 37
(20) is striking, although η(s, 1
2)is an independent transcendent of ζ(s).Bytheway,
the above reflection formula (21)forL(s) was also obtained by Oscar Schlömilch;
in 1849 he presented it as an exercise for students [55], and then, in 1858, he pub-
lished the proof [56]. Yet, it should be recalled that an analog of formula (20)forthe
alternating ζ-function η(s) and formula (21)forL(s) were already conjectured in
1749 by Euler, see pp. 94 and 105 of [20] respectively. However, Euler’s results are
usually not considered as rigorous proofs. Euler, first, studied the ratio η(1−n)/η(n)
for n=1,2,...,10. Then, by the method of mathematical induction, he conjectured
that in general
η(1−n)
η(n) =−(2n−1)(n −1)!
(2n−1−1)πncos πn
2,n=1,2,3,... (22)
Next, Euler showed that this formula remains valid for negative nas well. Finally, he
verified it analytically for n=1
2and numerically for n=3
2.15 As regards the function
L(s), Euler contented himself with the statement of the reflection formula (21) and
added that it can be derived analogously. By the way, for Euler, formula (22)was
not only interesting in itself, but was also a means by which could probably help in
the closed-form evaluation of η(n) for n=3,5,7,... But since η(1−n) for such
nvanishes and so does cos 1
2πn, he faced (after the performance using l’Hôpital’s
rule) a more difficult series (−1)k−1kn−1lnk,k1. Note that the main advantage
of Euler’s (22) and Malmsten’s (21) reflection formulas is that they can be verified
numerically because in each both sides converge for 0 <s<1, while Riemann’s
formula (20) requires the notion of analytic continuation.
Malmsten and colleagues also studied integrals containing an arctangent with hy-
perbolic functions. These studies resulted in an interesting relationship between the
Γ-function and the Ψ-function of a rational argument:
Ψm
n=−γ−ln2πn−π
2ctg πm
n−2
n−1
l=1
cos 2πml
n·lnΓl
n,m<n(23)
where mand nare positive integers [40, p. 57, and see also pp. 56 and 72]. Malmsten
and colleagues noticed that this relationship had not received sufficiently attention of
mathematicians. On the one hand, the reader may remark that the right part of (23)
may be easily transformed into elementary functions with the help of the reflection
formula for the Γ-function. The pairwise summation of all terms in the sum over
l(the first term with the last one, the second term with that before last, and so on)
makes the Γ-functions in the right part totally vanish:
al+an−l=cos 2πml
n·lnΓl
n+lnΓ1−l
n
=cos 2πml
n·lnπ−ln sin πl
n,(24)
15A sketch of the Euler’s proof of formula (22) may be also found in [31, pp. 23–26].
38 I.V. Blagouchine
where aldenotes the lth term of the aforementioned sum.16 In the case in which n
is even, there will be one term which will not be concerned by the above simplifi-
cation: the middle term ln Γ(1/2), but its value is well known and does not provide
any additional information about the Γ-function. Consequently, formula (23) may be
regarded as a variant of Gauss’ digamma theorem.17 On the other hand, (23) repre-
sents also a kind of discrete cosine transform (and more generally, a kind of finite
Fourier transform), which have a huge quantity of applications in engineering, espe-
cially in signal processing and related disciplines. From this point of view, Malmsten
was right when saying that these formulas were not sufficiently studied. In our work,
we will derive many similar formulas (see exercises no. 3,6,9,13,11,14,25,41,46,
48,49 in Sect. 4). In particular, Malmsten’s integrals of the first order18 can be often
expanded in such a kind of finite Fourier series having logarithm of the Γ-function as
coefficients. As a consequence, inverse transform may be also derived; the latter, inter
alia, provides values of the Γ-function at rational argument as functions of Malm-
sten’s integrals (see exercises no. 58–61). We will also give Parseval’s theorem for
such expansions (exercise no. 62). In addition, we will show that similar connections
exist also for higher polygamma functions (exercise no. 48-b), as well as for Stieltjes
constants (exercises no. 63–67).
Finally, we remark that in [40] some quantity of results were derived by means of
divergent series,19 but they were later re-obtained by Malmsten [41] by using other
methods.
2.3 Malmsten’s results and the Gradshteyn and Ryzhik’s tables
The famous Gradshteyn and Ryzhik’s tables [28] contains more than 20 formulas due
to Malmsten and his colleagues. They were borrowed via Bierens de Haan’s tables
[62] and [61]. These are, for example, formulas no. 4.325-3,4,5,6,7,8,9, both formulas
in no. 4.332, all formulas in no. 4.371 and 4.372, first three formulas in no. 4.373,
formula no. 4.267-3 and some others. By the way, several errors crept into no. 4.372:
in no. 4.372-1 and 4.372-2, the lower bound of both integrals should be 0 instead
of 1. The same errors appear in Prudnikov et al.’s tables no. 2.6.29-1 and no. 2.6.29-2
[53, vol. I].20 Also, the upper bound of the sum for the case “m+nis even” 1
2(n −1)
should be replaced by the integer part 1
2(n −1)(because when m+nis even, n
is not necessarily odd). Moreover, one should also add m<n(otherwise, integrals
on the left diverge). In no. 4.325-7, as showed in Malmsten’s et al. works, parameter
tshould be in the range (−π,+π). This is because the integral on the left is 2π-
periodic, while the right part is not periodic; both parts coincide only if t∈(−π, +π).
16We performed similar simplification for the Ψ-function in exercise no. 11,formula(48).
17Gauss presented the proof of this theorem in January 1812 [26].
18By the order of Malmsten’s integral we mean the order of poles of the corresponding integrands.
19The use of divergent series was especially common in the 18th century, see, for instance, the excellent
monograph [31].
20One of these two errors comes from the Bierens de Haan’s tables (latter borrowed them in part from
misprints in Malmsten’s work [41]; Bierens de Haan even complained about the number of misprints in
this work, see [61, p. 265]). For example, integrals’ bounds are incorrect in [62, Table 148-1,2,3,4], [61,
Table 191-1,2,3,4,5,6], [41, Eq. (10), (12) for both integrals]. The reader should be also careful with these
sources since integrands in Gradshteyn and Ryzhik’s tables are presented in other form.
Malmsten’s integrals and their evaluation by contour integration 39
Another error crept into no. 4.332-2. This integral is a rewritten version of (2a), and
hence, its right-hand side must be exactly as in (2a), i.e. √2πshould be replaced
by 3
√2π(this misprint was previously pointed out by the authors of [44]). All these
errors are present in the last 7th edition of the Gradshteyn and Ryzhik tables and in
all other editions, including the original Russian edition.
3 The proposed method
3.1 Preliminaries
Malmsten and colleagues used for their derivations elementary functions and the Γ-
function, without resorting to other special functions. The proposed method neither
uses special functions, except the Γ-function, and is based on the theory of functions
of a complex variable, and more precisely, on the contour integration technique. On
the one hand, such a method turns out to be much more straightforward than Malm-
sten’s method and, at the same time, it has wider applicability. On the other hand, it is
simpler than most modern methods which tend to resort to “heavy” special functions.
Moreover, the proposed method allows one to evaluate not only Malmsten’s integrals
from [40] and [41], but also many others (such results are given in Sect. 4).
The theory of functions of a complex variable, also known as complex analysis,
is one of the most beautiful and useful branch of mathematics having many versa-
tile applications. One such applications is the evaluation of integrals by means of
Cauchy’s residue theorem. This application is also known as the contour integration
method. Cauchy’s residue theorem states that for the function f(z)—which is ana-
lytic and single-valued inside and on a simple closed curve Lexcept possibly for a
finite number of isolated singularities—the contour integral
‰
L
f(z)dz=⎧
⎪
⎪
⎨
⎪
⎪
⎩
0,if f(z) has no singularities inside L,
2πi
m
l=1
res
z=zl
f(z), otherwise,
(25)
where {zl}m
l=1are the isolated singularities of the function f(z) enclosed by the con-
tour L. Technically, the application of the residue theorem for the evaluation of a
given integral is done in two stages. First, by decomposition of the integration path L
the line integral on the left in (25) is reduced to the evaluated integral (it can be done
in many different ways; often, the latter appears as by-product). Then, one calculates
the sum of the residues on the right in (25). The final result is obtained by equating
both parts of (25). Numerous examples of such evaluations may be found in classical
complex analysis literature [3,16,22,23,25,32,34,42,43,59,60,69], [58, vol. III,
part 2], [33,65].
A major difficulty encountered when evaluating logarithmic integrals21 by
Cauchy’s residue theorem consists in the following trade-off. On the one hand, the
logarithm, being a typical multiple-valued function, has branch points, which should
21We will not consider here indirect methods, such as, for example, evaluation of logarithmic integrals
based on the differentiation of the integrand (which does not contain a logarithm) with respect to a param-
eter.
40 I.V. Blagouchine
not lie within the integration contour. On the other hand, not any integration path will
lead to the wanted integral. In simple cases, e.g.,
∞
ˆ
0
R(x)lnnxdx,
∞
ˆ
0
R(x)lnx2+a2dx,
∞
ˆ
0
R(x)
ln2x+π2dx,
where a>0, n∈Nand R(x) denotes an arbitrary rational function of x(eveninthe
first two cases), the evaluation may be succeed by considering respectively
‰R(z)lnn(z) d z, ‰R(z)ln(z +ia)dz, ‰R(z)
lnz−πi dz,
taken around simple integration contours (e.g., semi-circle, circle with a cut) with
the help of Jordan’s lemma, see e.g. [59, pp. 187–188, 193–194, 197–198],22 [69,
pp. 129–132], [22, chapter VI, § 3], [23, pp. 281–296]. But the evaluation of more
complicated logarithmic integrals may become a difficult task, because it may be
very hard (or even impossible) to find an appropriate line integral. A typical example
of such logarithmic integrals is the simplest Malmsten’s integral (1). Consider, for
example, its hyperbolic form. On the one hand, Jordan’s lemma cannot be applied
to this integral (it is sufficient to note that the integrand remains unbounded on the
imaginary axis). On the other hand, the integrand has infinitely many poles in both
semi-circles, which leads to an infinite series on the right part of (25). As regards
the ln ln-form of Malmsten’s integral (1), the integrand fulfills conditions of Jordan’s
lemma in both half-planes, but on the one hand, it has two branch points, 0 and 1,
which should be properly indented, and on the other hand, the integral over (−∞,0]
can be hardly reduced to that over [0,+∞). Thus, in order to evaluate such kinds of
integrals, one may be led to consider unusual integration paths and more sophisticated
forms of the integrand.
3.2 Introduction
Consider the following general family of logarithmic integrals:
+∞
ˆ
−∞
Rexlnx2+a2dx, a ∈R,(26)
where R(·)denotes a rational function, and its particular case a=0
+∞
ˆ
−∞
Rexln|x|dx =∞
ˆ
1
R(u) +R(u−1)
ulnln udu=
1
ˆ
0
R(y) +R(y−1)
ylnln 1
ydy ,
22The readers of this book should beware of misprints and of some incorrect results. Answers in exercises
no. 84, 88, 91 are incorrect; on the p. 189, 2πi is forgotten in the right part of equation (1). Several errors
were corrected in the recent second edition of this book, but the few ones are still present, e.g. answer in
no. 7.91 is incorrect, the above-mentioned coefficient 2πi is absent.
Malmsten’s integrals and their evaluation by contour integration 41
where we made two consecutive changes of variable x=lnuand u=y−1. Denoting
for brevity
Q(u) ≡1
uR(u) +Ru−1 (27)
the last line takes the form
+∞
ˆ
−∞
Rexln|x|dx =∞
ˆ
1
Q(u) ln ln udu=
1
ˆ
0
Q(y) ln ln 1
ydy. (28)
Thus, for any integral of the form (28), we may formulate the following statement:
if it is possible to find such a rational function R(u) that the function Q(u) may
be represented in the form (27), then last two integrals in (28)may be evaluated
via integral (26). The latter, as we come to see later, may be always expressed in
terms of the Γ-function and its logarithmic derivatives. Equation (27) implies also
that function Q(u) obeys the following functional relationship: Q(u−1)=u2Q(u).
One of the consequences of this property is that the integrand (without lnln part)
remains unchanged when bounds [1,∞)are replaced with [0,1]. Consequently, the
above statement may be reformulated as follows: if the integrand from (28)(that of
the integral in the middle) without lnln-part, after a change of variable u=y−1,
remains invariant and only bounds [1,∞)are replaced with [0,1],i.e.if
∞
ˆ
1
Q(u) ln ln udu=
1
ˆ
0
Q(y) ln ln 1
ydy,
then it can be always evaluated by the proposed method, and the result may be ex-
pressed in terms of the Γ-function and its logarithmic derivatives.
Now, one can remark that all ln ln-integrals that Malmsten and colleagues evalu-
ated in [40] and [41] are particular cases of integrals (28) and fulfill the above condi-
tion on Q(u). For example, the simplest Malmsten’s integral (1) is obtained by tak-
ing R(ex)=1
4ch−1x, which gives Q(u) =1/(1+u2). By the way, it is curious that
Malmsten did not notice that integrands of all his integrals obey Q(u−1)=u2Q(u).
It is also obvious that if Q(u−1)=u2Q(u), then
∞
ˆ
1
Q(u) ln ln udu=
1
ˆ
0
Q(y) ln ln 1
ydy.
For instance, the following integral
∞
ˆ
1
lnln u
1+udu =
1
ˆ
0
lnln 1
y
1+ydy =∞
ˆ
1
lnln u
u2+udu =∞
ˆ
0
lnx
ex+1dx =−1
2ln22
which appears as no. 4.325-1 in [28], cannot be evaluated by the proposed method,
at least directly. However, such kind of integrals can be often evaluated by other
methods. For example, in some cases, the series expansions method turns out to be
42 I.V. Blagouchine
very useful for this aim, see e.g. exercises no. 18–19 in Sect. 4. There are also other
methods that merit being mentioned in this context, but most of them resort to higher
transcendences than the Γ-function. For instance, Adamchik [2] used the Hurwitz
ζ-function and showed that if Q(y) in (28) is a cyclotomic polynomial, then the cor-
responding integral may be always expressed in terms of derivatives of the Hurwitz
ζ-function. Vardi [67], Vilceanu [68] and Bassett [8] suggested the use of the Dirich-
let L-function. As regards Medina et al. [44], the authors evaluated these integrals
with the help of polylogarithms.
3.3 The method for the evaluation of logarithmic and arctangent integrals
The Γ-function
Γ(z)=∞
ˆ
0
tz−1e−zdt
is one of the oldest special functions which was introduced in analysis by Leon-
hard Euler. It was studied in detail by Euler himself, as well as by many other great
mathematicians such as Carl Friedrich Gauss, Adrien-Marie Legendre, Karl Weier-
strass, Otto Hölder and many others.23 Classically, the Γ-function was defined only
for positive values of its argument z, but it is now well known that it can be ana-
lytically continued to the entire complex plane except for simple poles at the points
0,−1,−2,.... In addition, since for each integer n,Γ(n)=(n −1)!=0, by virtue
of the reflection formula, one may easily establish that Γ-function has no zeros at all.
As a consequence, ln Γ(z)has no branch points. From the recurrence relationship for
the Γ-function Γ(z+1)=zΓ(z), one can easily deduce an analogous functional
relationship for the logarithm of the Γ-function, denoted Λ(z) for brevity:
Λ(z +1)−Λ(z) =ln z(29)
It is therefore apparent that the use of the logarithm of the Γ-function may lead to
the appearance of the logarithm. But why should one chose the logarithm of the Γ-
function rather than simply the logarithm? The main advantage of the Λ(z) function
over the logarithm is that the former has no branch points at all, which allows one to
use Cauchy’s residue theorem with much less restriction to the choice of the integra-
tion contour.
Let R(·)be a real rational function of ex,x∈R, such that for some β>0, we
always have
βlnβ·max
ϕ∈[0,2π]Re±β+iϕ→0asβ→∞.(30)
Consider now the line integral
‰
Lβ
RezΛz
2πi +αdz, α 0
23An interesting and quite well-written historical overview on the Γ-function is given in [19]. Motivated
readers who are not afraid of French and German are also invited to take the look at these classic books
[15,27,49]and[6].
Malmsten’s integrals and their evaluation by contour integration 43
taken around a rectangle with vertices at [(−β,0), (+β,0), (+β, 2πi),(−β,2πi)]
designated by Lβ. Bearing in mind that the positive direction is counterclockwise,
the above contour integral may be split in four integrals as follows:
‰
Lβ
RezΛz
2πi +αdz =
+β
ˆ
−β
... dz+
β+2πi
ˆ
β
... dz+
−β+2πi
ˆ
β+2πi
... dz+
−β
ˆ
−β+2πi
... dz
(31)
where the integrands on the right were omitted for brevity. Now let β→∞. Taking
all necessary precautions, the last equation becomes
‰
L∞
RezΛz
2πi +αdz
=lim
β→∞⎧
⎪
⎨
⎪
⎩
+β
ˆ
−β
... dz +
−β+2πi
ˆ
β+2πi
... dz
⎫
⎪
⎬
⎪
⎭+lim
β→∞
β+2πi
ˆ
β
... dz +lim
β→∞
−β
ˆ
−β+2πi
... dz (32)
Now, if (30) holds, then it can be shown that first limit in (32) converges to some
finite non-zero quantity, while second and third limits equal zero. The latter may be
proved in the following manner. We first notice that the behavior of the logarithm of
the Γ-function in the sector |argz|<π/2 when |z|→∞may be described by the
following asymptotic formula24
Λ(z) =z−1
2lnz−z+1
2ln2π+2∞
ˆ
0
arctg(x/z)
e2πx −1dx
O(z−1)
=zlnz+O(z), (33)
see e.g. [42, vol. II, pp. 315–321], [39, pp. 88–89], [22, chapter VI, § 6], [1, no.
6.1.40–6.1.41, 6.1.43–6.1.44, 6.1.50], [53, vol. I, no. 2.7.5-6], [26, p. 33]. Hence, the
integral in the second limit on the right in (32), after a change of variable z=β+iϕ,
may be estimated, for sufficiently large β, in the following manner:
β+2πi
ˆ
β
RezΛz
2πi +αdz
2πmax
ϕ∈[0,2π]Reβ+iϕΛβ+iϕ
2πi +α
2π·max
ϕ∈[0,2π]Λβ+iϕ
2πi +α·max
ϕ∈[0,2π]Reβ+iϕ
∼βln β
2π·max
ϕ∈[0,2π]Reβ+iϕ→0asβ→∞
24The non-asymptotic part of this formula (that containing an infinite integral) is also known as the second
Binet’s expression for the logarithm of the Γ-function [12, pp. 335–336], [71, pp. 250–251], [9,vol.I,
p. 22, Eq. 1.9(9)] (for more details, see also exercise no. 40 in the last section of this manuscript). As
regards its asymptotic form, it was already known to Gauss [26, p. 33], and in a more simple form (for
natural z), to Euler [21, part II, Chap. VI, p. 466], to Stirling and to de Moivre.
44 I.V. Blagouchine
thanks to condition (30). In like manner, we can show that the third limit in (32)
vanishes also. Consider now the second integral in the right part of (32). By making a
change of variable z=x+2πi and by noticing that R(ez)is a 2πi-periodic function,
this integral reduces to
−β+2πi
ˆ
β+2πi
RezΛz
2πi +αdz =−
+β
ˆ
−β
RexΛx
2πi +α+1dx.
Equation (32) may be therefore rewritten as
‰
L∞
RezΛz
2πi +αdz
=lim
β→∞
+β
ˆ
−β
RexΛx
2πi +α−Λx
2πi +α+1dx
=−+∞
ˆ
−∞
Rexlnx
2πi +αdx
=−+∞
ˆ
−∞
Rexln(x +2πiα)dx +ln2π+πi
2·+∞
ˆ
−∞
Rexdx
JR
.(34)
We note, in passing, that because condition (30) holds the first integral in the second
line converges, and so does the integral JR.
On the other hand, the left part of the last equation may be computed by the residue
theorem
‰
L∞
RezΛz
2πi +αdz (35)
=2πim
l=1
res
z=zlRezΛz
2πi +α+1
2m
l=1
res
z=˜zlRezΛz
2πi +α,
where {zl}m
l=1are the isolated singularities of the integrand lying within the strip
0<Im z<2π, and {˜zl}m
l=1are those whose imaginary part is exactly 0 or 2π(i.e.,
they lie on the integration path). By equating right-hand sides of (34) and of (35), we
Malmsten’s integrals and their evaluation by contour integration 45
get
+∞
ˆ
−∞
Rexln(x +2πiα)dx =JRln2π+πi
2(36)
−2πim
l=1
res
z=zlRezΛz
2πi +α+1
2m
l=1
res
z=˜zlRezΛz
2πi +α
Now, on taking real parts, we obtain
+∞
ˆ
−∞
Rexlnx2+4π2α2dx =2JRln2π
+4πImm
l=1
res
z=zlRezΛz
2πi +α+1
2m
l=1
res
z=˜zlRezΛz
2πi +α
while equating imaginary parts yields
+∞
ˆ
−∞
Rexarctg2πα
xdx =π
2
∞
ˆ
0Rex−Re−xdx
−2πRem
l=1
res
z=zlRezΛz
2πi +α+1
2m
l=1
res
z=˜zlRezΛz
2πi +α
Rewriting the first equation with α=a
2πand the second one with α=1
2πa , and
recalling that arctg 1
x=π
2sgnx−arctg xfor any real xexcept zero, we arrive at
integral (26)
+∞
ˆ
−∞
Rexlnx2+a2dx =2JRln 2π(37)
+4πImm
l=1
res
z=zlRezlnΓz+ai
2πi +1
2m
l=1
res
z=˜zlRezln Γz+ai
2πi
and at another integral:
+∞
ˆ
−∞
Rexarctg(ax ) dx (38)
=2πRem
l=1
res
z=zlRezlnΓza +i
2πai +1
2m
l=1
res
z=˜zlRezln Γza +i
2πai .
46 I.V. Blagouchine
For the logarithmic integral, the particular case a=0 may be of special interest.
From (37), we easily get
+∞
ˆ
−∞
Rexln|x|dx =JRln 2π(39)
+2πImm
l=1
res
z=zlRezlnΓz
2πi+1
2m
l=1
res
z=˜zlRezln Γz
2πi.
In contrast, for the arctangent integral, the limiting case a→∞reveals to be more
interesting. The fact that lim
a→∞arctg ax =π
2sgnxgives the opportunity to evaluate
integrals of some odd functions over interval [0,∞). Making a→∞, we obtain
from (38)
∞
ˆ
0Rex−Re−xdx (40)
=4Re
m
l=1
res
z=zlRezlnΓz
2πi+1
2m
l=1
res
z=˜zlRezln Γz
2πi.
If R(ex)is odd then R(ex)−R(e−x)=2R(ex), while if R(ex)is even the last integral
vanishes identically.
Formulas (37)–(40) allow one to compute many kind of different integrals con-
taining logarithms, inverse trigonometric functions and many others. For example,
these integrals
+∞
ˆ
−∞
R(ex)
(x2+a2)ndx,
+∞
ˆ
−∞
xR(e
x)
(x2+b2)ndx, b ≡a−1,n∈N,
may be straightforwardly obtained from derived ones by a simple differentiation with
respect to the parameter a. In addition, the evenness or oddness of R(ex)may simply
calculations. Moreover, if the integral on the left diverges in the classical sense, in
some cases, it can be still evaluated in the sense of the Cauchy principal value with
the help of the above formulas. To illustrate these matters more vividly, the next
section provides several beautiful examples of applications.
As to the integral JR, it is difficult to treat the general case, so each integral must
be considered individually. For example, it may be simply an elementary integral.
Otherwise, its computation may be performed by the contour integration via
‰Rezdz
taken around an infinitely long rectangle of breadth πor 2πby the method analogous
to that in (31). Such integrals are also exhaustively treated in [59, pp. 186, 197–198],
[69, p. 132], [23, pp. 276–277], [25].
Malmsten’s integrals and their evaluation by contour integration 47
At the end of the demonstration, it may be of interest to remark that some quan-
tity of exercises considering similar methods of contour integration involving the
Γ-function and its derivatives are given in the excellent Russian book [23], which,
unfortunately, was never translated into other languages, and for which reason it prob-
ably remains inaccessible to most readers. It seems also quite fair to say that the above
exposition is much inspired from exercises no. 31.30–31.32. In exercise no. 31.32, the
reader is asked to prove a simpler variant of formula (37) provided R(ex)∈L1.Re-
grettably, despite the high complexity of problems (higher than in the well-known
collections [59] and [69]), this book does not contain solutions, nor even hints, only
answers are provided at the end of each section.
3.4 Application to Malmsten’s integrals
3.4.1 The simplest Malmsten’s integral
Let’s calculate Malmsten’s integral (1) by means of formula (37). Since the function
R=ch−1x, being a rational function of ex, it easily satisfies (30). Besides, it is a
meromorphic function having two simple poles within the strip 0 Im z2πat
points πi/2 and 3πi/2. Taking additionally into account that ch−1xis even, (37)
gives
∞
ˆ
0
ln(x2+a2)
chxdx =2πImres
z=1
2πi
lnΓ(z+ai
2πi )
chz+res
z=3
2πi
lnΓ(z+ai
2πi )
chz+ln(2π)
+∞
ˆ
−∞
dx
chx
π
=2πlnΓ3
4+a
2π−lnΓ1
4+a
2π+πln 2π. (41)
In the last line we may easily recognize Malmsten’s general formula (14). Letting
a→0, we arrive at (1) as follows:
1
2
∞
ˆ
0
lnx
chxdx =1
4lim
a→0
∞
ˆ
0
ln(x2+a2)
chxdx =π
2lnΓ3
4−lnΓ1
4+π
4ln2π
=π
2lnΓ(3/4)
Γ(1/4)√2π=π
2ln 2 +3π
4lnπ−πln Γ1
4,(42)
where the final simplification is done with the help of the reflection formula for the Γ-
function. The final result is, therefore, completely expressed in terms of mathematical
constants.
3.4.2 Other Malmsten’s integrals
Consider now another Malmsten’s integral, mentioned by Vardi and others re-
searchers as well, namely the integral (2b). After changes of variable y=u−1,
48 I.V. Blagouchine
u=ex, it can be rewritten as
1
ˆ
0
lnln 1
y
1−y+y2dy =∞
ˆ
1
lnln u
1−u+u2du =∞
ˆ
0
lnx
2chx−1dx
Poles of the latter integrand in the strip 0 Im z2πare simple and occur at points
πi/3 and 5πi/3. Application of formula (37) yields
∞
ˆ
0
ln(x2+a2)
2chx−1dx =2πImres
z=1
3πi
lnΓ(z+ai
2πi )
2chz−1+res
z=5
3πi
lnΓ(z+ai
2πi )
2chz−1(43)
+ln(2π)
+∞
ˆ
−∞
dx
2chx−1
4π
3√3
=2π
√3lnΓ5
6+a
2π−lnΓ1
6+a
2π+2ln2π
3
Letting a→0 yields
∞
ˆ
0
lnx
2chx−1dx =1
2lim
a→0
∞
ˆ
0
ln(x2+a2)
2chx−1dx
=π
√3lnΓ5
6−lnΓ1
6+2ln2π
3
=π
√3lnΓ(5/6)
Γ(1/6)
3
4π2=2π
√35
6ln2π−ln Γ1
6,
which is identical with (2b). Again, at the last stage, the reflection formula was em-
ployed. Finally, in view of the fact that Γ(1/6)=31
22−1
3π−1
2Γ2(1/3), see e.g. [15,
p. 31], the latter formula may be written in terms of Γ(1/3), and hence
1
ˆ
0
lnln 1
y
1−y+y2dy =∞
ˆ
1
ln ln u
1−u+u2du =∞
ˆ
0
lnx
2chx−1dx
=π
3√37ln2+8lnπ−3ln3−12 ln Γ1
3.(44)
Analogously, it can be shown that
1
ˆ
0
lnln 1
y
1+y+y2dy =∞
ˆ
1
ln ln u
1+u+u2du =∞
ˆ
0
lnx
2chx+1dx
=π
6√38ln2π−3ln3−12ln Γ1
3.(45)
Malmsten’s integrals and their evaluation by contour integration 49
The unique difference between integrals (44) and (45) is the location of poles of inte-
grands. For the former, they occur(for the hyperbolic form) in the strip 0 Im z2π
at πi/3 and 5πi/3, while for the latter they occur at 2πi/3 and 4πi/3.
Another frequently encountered Malmsten’s integral (however, not mentioned by
Vardi) is this one:
1
ˆ
0
lnln 1
y
(1+y)2dy =∞
ˆ
1
ln ln u
(1+u)2du =1
4
∞
ˆ
0
lnx
ch21
2xdx =1
2
∞
ˆ
0
lnx
chx+1dx (46)
see e.g. [41,p.24],[62, Table 147-7, 257-4], [61, Table 190-7], [28, no. 4.325-3]. It
can be also evaluated by the proposed method. In the strip 0 Im z2πthe inte-
grand has one double pole at z=πi. Proceeding as above, we find that
∞
ˆ
0
ln(x2+a2)
chx+1dx =2πImres
z=πi
lnΓ(z+ai
2πi )
chz+1+ln(2π)
+∞
ˆ
−∞
dx
chx+1
2
=2Ψ1
2+a
2π+ln2π.(47)
When atends to zero, we have
1
2
∞
ˆ
0
lnx
chx+1dx =1
4lim
a→0
∞
ˆ
0
ln(x2+a2)
chx+1dx =1
2Ψ1
2+ln2π
=1
2−γ+ln π
2,
which completes the evaluation of (46). Other Malmsten’s integrals from [40] and
[41] may be evaluated similarly.
4 New results, problems and exercises
This section is designed as a collection of original exercises to be worked out by the
readers. Exercises marked with an ∗contain new results which were never, to our
knowledge, released before (except if otherwise stated). There are also some known
results which were historically obtained by other methods. For such problems original
sources of the formulas are provided. The results are presented in a quite general
form, which is why most of the formulas contain different parameters with respect to
which they can be differentiated or integrated.
50 I.V. Blagouchine
4.1 Logarithmic integrals containing hyperbolic functions
4.1.1 Main results obtained by the contour integration method
1By using formula (37), verify that for any a0 and Re b>0
∞
ˆ
0
ln(x2+a2)
chbx dx =2π
blnΓ3
4+ab
2π−lnΓ1
4+ab
2π+1
2ln 2π
b
Hint: Make a suitable change of variable in (41).
Nota bene: This formula for b>0 can be found in Gradshteyn and Ryzhik’s tables
[28, no. 4.373-1].
2
*Prove by the contour integration method the following formula
∞
ˆ
0
ln(x2+a2)
chbx +cos ϕdx =2π
bsinϕln Γ1
2+ab +ϕ
2π−lnΓ1
2+ab −ϕ
2π+ϕ
πln 2π
b
a0, Reb>0, |Re ϕ|<π,ϕ= 0; for ϕ=0, see formula (47) and exercise no. 6.
Note that this formula remains valid even for complex values of ϕ.Ifϕis imaginary
pure, then ϕ=it,t∈R, and the denominator of the integral takes the form ch x+cht.
Such integrals can be always evaluated via the above formula. Moreover, even if ϕ
lies outside the vertical strip |Re ϕ|<π, the integral can be still computed in the
sense of the Cauchy principal value. Note also that in the particular case ϕ=±π/2,
the above formula reduces to that obtained in the previous exercise.
Hint: By considering
‰
0Im z2π
e−iαz
chz+cos ϕdz, prove first
+∞
ˆ
−∞
cosαx
chx+cos ϕdx =2π
sin ϕ·sh ϕα
shπα,|Im α|<1,
|Re ϕ|<π, ϕ= 0.
Then, let α→0. Note that in exercise no. 28.19-4 [23], where the last integral also
appears, there is a slight inaccuracy concerning the domain of convergence: it is de-
fined only for Im ϕ>0.
Nota bene:Malmsten[41, p. 24] discovered a particular case of this formula for
a=0, b=1 and real ϕ∈(−π, +π). Formula (63) on the p. 24 in [41] is actually a
rewritten version of such a particular case (see also [61, Table 274-12 ⇒Table 190-
9], [62, Table 257-7 ⇒Table 147-9], [28, no. 4.371-2]). The same particular case was
independently rediscovered by Medina et al. in [44]. As regards the general formula
given above, it seems to be not new.
Malmsten’s integrals and their evaluation by contour integration 51
3
*By the contour integration method, prove that if pis a discrete parameter chosen
so that p=bm/n, where Re b>0 and numbers mand nare positive integers such
that m<n, then for any a0, one always has
(a) ∞
ˆ
0
sh px ·ln(x2+a2)
sh bx dx =π
btg mπ
2n·ln 2πn
b
+2π
b
2n−1
l=1
(−1)lsin mπl
n·lnΓl
2n+ab
2πn
(b) ∞
ˆ
0
chpx ·ln(x2+a2)
chbx dx =π
bsec mπ
2n·ln 2πn
b
−2π
b
2n−1
l=0
(−1)lcos (2l+1)mπ
2n·lnΓ2l+1
4n+ab
2πn
(c) ∞
ˆ
0
chpx ·ln(x2+a2)
chbx +cos ϕdx =2π
bsin mϕ
n·cscϕ·csc mπ
n·ln 2πn
b
+2π
bsinϕ
n−1
l=0cos (2l+1)mπ +mϕ
n·lnΓ2l+1
2n+ab +ϕ
2πn
−cos (2l+1)mπ −mϕ
n·lnΓ2l+1
2n+ab −ϕ
2πn
(d) ∞
ˆ
0
chpx ·ln(x2+a2)
chbx +1dx =2πm
bn csc mπ
n·ln 2πn
b
−4πm
bn
n−1
l=0
sin (2l+1)mπ
n·lnΓ2l+1
2n+ab
2πn
+2
bn
n−1
l=0
cos (2l+1)mπ
n·Ψ2l+1
2n+ab
2πn
where in (c) |Re ϕ|<π,ϕ= 0; for ϕ=0, see (d). For continuous and complex
values of p, see no. 63–64.
Hint: As regards exercise (a), first, put for simplicity b=1 and rewrite the integral
for x=yn as follows:
∞
ˆ
0
sh(mx/n) ·ln(x2+a2)
sh xdx =n
∞
ˆ
0
sh my ·ln(y2n2+a2)
sh ny dy
=2nlnn
∞
ˆ
0
sh my
sh ny dy +n
∞
ˆ
0
sh my ·ln(y2+(a/n)2)
sh ny dy.
52 I.V. Blagouchine
Then, by noticing that the integrand of the last integral on the right is a rational func-
tion of eyand, hence, is 2πi-periodic, apply formula (38). When evaluating residues,
do not forget that function sh mz
sh nz has removable singularities at z=0, z=πi,z=
2πi and poles of the first order at z=iπl/n,l=1,2,3,...,n−1,n+1,...,2n−1.
Integrals (b)–(d) are evaluated similarly.
Nota bene: Integrals (a) and (b) for b=π, as we explained in the first part of our
work, were evaluated by Malmsten and colleagues, see expressions (13) and (15),
respectively. Their formulas slightly differ from ours because they separately con-
sidered cases m+nodd and m+neven. Besides, both formulas (13) and (15) can
be further simplified. Integrals (c) and (d) seem to be not evaluated previously. For-
mula (c) is an important general formula from which several particular cases may be
derived via an appropriate limiting procedure. For example, integral d), as well as
formulas in exercises no. 6b–d may be obtained in this way.
4
*Prove by the contour integration method the following general formulas
(a) p.v.
+∞
ˆ
−∞
ln(x2+a2)
sh x±sh tdx =±2π
chtπ
2−arctg a
t±4tln2π
cht
±4π
chtImln Γa
2π+it
2π−lnΓ1
2+a
2π−it
2π
(b) +∞
ˆ
−∞
ln(x2+1−t2)
sh x±sh tdx =±2π
chtπ
2−arctg √1−t2
t±4tln2π
cht
±4π
chtImln Γ√1−t2
2π+it
2π−lnΓ1
2+√1−t2
2π−it
2π
By the way, formula (b) may be also written in a slightly different form:
(b)
∞
ˆ
0
ln(x2+1−t2)
sh2t−sh2xdx =2π
sh 2tπ
2−arctg √1−t2
t+4tln2π
sh 2t
+4π
sh2tImln Γ√1−t2
2π+it
2π−lnΓ1
2+√1−t2
2π−it
2π
where in (a) a0, t>0, and in (b) and (b)0<t1.
Hint: By considering
‰
0Im z2π
e−iαz
sh z+shtdz, prove first
Malmsten’s integrals and their evaluation by contour integration 53
p.v.
+∞
ˆ
−∞
e−iαx
sh x+sh tdx =πi
cht·sh απ e−iαt −eiαt ch απ,
provided |Im α|<1 and −∞<t<∞. Then, let α→0.25
Nota bene: In spite of the fact that some integrals in this exercise may be evaluated
only in the sense of the Cauchy principal value, their evaluation is not as useless as it
may first appear. In fact, by an appropriate choice of the numerator of the integrand,
it is often possible to get rid of the p.v.sign. For instance, with the help of the last
formula, we may arrive at these beautiful and quite non-trivial26 convergent integrals
(c) +∞
ˆ
−∞
x+t
sh x+sh tdx =2∞
ˆ
0
xsh x−tsht
sh2x−sh2tdx =π2+4t2
2cht,
(d) +∞
ˆ
−∞
x2−t2
sh x+sh tdx =−t(π2+4t2)
3cht,
(e) ∞
ˆ
0
x2−t2
sh2x−sh2tdx =t(π2+4t2)
3sh2t,
(f) +∞
ˆ
−∞
sin αx +sin αt
sh x+sh tdx =2∞
ˆ
0
sin αx ·sh x−sinαt ·sh t
sh2x−sh2tdx
=1
cht2tsin αt +πthαπ
2cosαt,
(g) +∞
ˆ
−∞
cosαx −cos αt
sh x+shtdx =1
chtπcthαπ
2sinαt −2tcos αt,
(h) ∞
ˆ
0
cosαx −cos αt
sh2x−sh2tdx =1
sh 2t2tcosαt −πcthαπ
2sin αt.
The above results hold for any t∈(−∞,+∞)and αlying within the strip
|Im α|<1. The reader is also asked to prove these formulas as an exercise.
25For the evaluation of the integral JRlazy readers may directly use formula 1.4.7-15 from [53,vol.I,
p. 148]. Nevertheless, it is highly recommended that readers employ the proposed method rather than the
ready formula, since the procedure for the calculation of the logarithmic integral is very similar.
26For instance, we have not found these integrals in [28], neither in [53].
54 I.V. Blagouchine
5
*By using results of the previous exercise, prove that
+∞
ˆ
−∞
ln|x|−ln t
shx−sh tdx =−2π
chtImln it
2π−lnΓ1
2−it
2π
−π2
2cht+2t
chtln t
2π
and
∞
ˆ
0
lnx−ln t
sh2x−sh2tdx =−2π
sh 2tImln Γit
2π−lnΓ1
2−it
2π−π2
2sh2t
+2t
sh 2tln t
2π
provided t>0.
6*By using Cauchy’s residue theorem, prove that for any a0, Reb>0 and
p=bm/n, numbers mand nbeing positive integers, the following formulas hold
(a) ∞
ˆ
0
ln(x2+a2)
ch2bx dx =2
bln π
b+Ψ1
2+ab
π.
If the product ab is a rational part of πor ais zero, then the above integral may be
always expressed in terms of elementary functions and Euler’s constant γ(in virtue
of Gauss’ digamma theorem). For example:
∞
ˆ
0
lnx
ch2xdx =
1
ˆ
0
lnarcth xdx =ln π
4−γ,
∞
ˆ
0
ln(x2+π2)
ch2xdx =2ln π
4+4−2γ
etc.
(b) ∞
ˆ
0
chpx ·ln(x2+a2)
ch2bx dx =πm
bn ·csc mπ
2n·ln 2πn
b
−2πm
bn
2n−1
l=0
sin (2l+1)mπ
2n·lnΓ2l+1
4n+ab
2πn
+1
bn
2n−1
l=0
cos (2l+1)mπ
2n·Ψ2l+1
4n+ab
2πn
Malmsten’s integrals and their evaluation by contour integration 55
(c) ∞
ˆ
0
ch2px ·ln(x2+a2)
ch2bx dx =πm
bn ·csc mπ
n·ln πn
b
−2πm
bn
n−1
l=0
sin (2l+1)mπ
n·lnΓ2l+1
2n+ab
πn
+1
bn
n−1
l=0
cos (2l+1)mπ
n·Ψ2l+1
2n+ab
πn+1
bln π
b+Ψ1
2+ab
π
(d) ∞
ˆ
0
sh2px ·ln(x2+a2)
ch2bx dx =πm
bn ·csc mπ
n·ln πn
b
−2πm
bn
n−1
l=0
sin (2l+1)mπ
n·lnΓ2l+1
2n+ab
πn
+1
bn
n−1
l=0
cos (2l+1)mπ
n·Ψ2l+1
2n+ab
πn−1
bln π
b+Ψ1
2+ab
π
where in (b) m<2n, and in (c)–(d) m<n.
Hint: For the integral (b), see exercise no. 3. As regards last the two integrals, they
may be obtained by a linear combination of integrals (a) and (b). In last two cases,
at the final stage, split both sums over l=0,1,2,...,2n−1 into two sums of equal
lengths and use duplication formulas in order to simplify the result.
7
*Show that for any a0 and Re b>0
(a) ∞
ˆ
0
ln(x2+a2)
2ch
2bx +1dx =πi
b√3ln Γ1
2+ab
π+ln(2+√3)
2πi
Γ1
2+ab
π−ln(2+√3)
2πi
+ln(2+√3)
b√3ln π
b
(b) ∞
ˆ
0
ln(x2+a2)
2ch
2bx −1dx =π
blnΓ3
4+ab
π−lnΓ1
4+ab
π+1
2ln π
b
Hint: In order to get formula (a), first use formula (37), then notice that sh ln(2±
√3)=±√3 and ln(2+√3)=−ln(2−√3). At the final stage, use the duplication
formula for the Γ-function.
Nota bene: The particular case of the integral (a) for a=0 and b=1 may be found
in the Prudnikov et al.’s tables [53]. However, the provided expression is completely
56 I.V. Blagouchine
different and its numerical verification (performed with the help of Maple 12 and
MATLAB 7.2) fails:
∞
ˆ
0
lnx
2ch
2x+1dx
−0.2686306939...
=√π
2√3
∞
n=1
(−1)nγ+ln4n
√nsin πn
3
−0.2977821762...
[53, vol. I, p. 534, no. 2.6.29-7]. A careful study of this formula shows that the error
consists in the misplaced square sign: ch2xshould be replaced by ch x2, that is to
say,
∞
ˆ
0
lnx
2chx2+1dx =√π
2√3
∞
n=1
(−1)nγ+ln4n
√nsin πn
3=−0.2977821762...
By the way, in a slightly different form the above formula appears in the so many
times cited Malmsten’s work [41, p. 15, Eq. (41)].
8
*Prove that for any a0 and Reb>0
(a) ∞
ˆ
0
ln(x2+a2)
ch2bx +sin2ϕdx =πi
bsin ϕsin2ϕ+1ln Γ1
2+ab
π+lnκ
2πi
Γ1
2+ab
π−lnκ
2πi
+lnκ
bsinϕsin2ϕ+1ln π
b,
where κ≡1+2sin2ϕ+2sinϕ sin2ϕ+1,ϕ∈C,
(b) ∞
ˆ
0
ln(x2+a2)
ch2bx −sin2ϕdx =2π
bsin2ϕln Γ1
2+ab+ϕ
π
Γ1
2+ab−ϕ
π
+4ϕ
bsin2ϕln π
b,⎧
⎪
⎨
⎪
⎩
|Re ϕ|<π
2,a=1,
|Re ϕ|π
2,a=1,
where the right-hand side should be regarded as a limit in cases ϕ=0 and ϕ=±π/2
(a=1). Note also that both integrals from exercise no. 7are actually particular cases
of above ones with ϕ=π/4.
9*Prove that for any a0, Re b>0 and p=bm/n, where numbers mand nare
positive integers such that m<2n,
(a) ∞
ˆ
0
chpx ·ln(x2+a2)
ch2bx +sin2ϕdx =π(κ m
2n−κ−m
2n)
2bsinϕsin2ϕ+1·csc mπ
2n·ln 2πn
b
Malmsten’s integrals and their evaluation by contour integration 57
−π
2bsinϕsin2ϕ+1Im
2n−1
l=0κm
2nlnΓ2l+1
4n+ab
2πn +lnκ
4πin
−κ−m
2nlnΓ2l+1
4n+ab
2πn −lnκ
4πin·exp (2l+1)mπi
2n
−κm
2nlnΓ2l+1
4n+ab
2πn −lnκ
4πin
−κ−m
2nlnΓ2l+1
4n+ab
2πn +lnκ
4πin·exp −(2l+1)mπi
2n,
where κ≡1+2sin
2ϕ+2sinϕsin2ϕ+1, ϕ∈R,
(b) ∞
ˆ
0
chpx ·ln(x2+a2)
ch2bx −sin2ϕdx =2π
bsin2ϕ·sin mϕ
n·csc mπ
2n·ln 2πn
b
+2π
bsin2ϕ
2n−1
l=0cos (2l+1)mπ +2mϕ
2n·lnΓ2l+1
4n+ab +ϕ
2πn
−cos (2l+1)mπ −2mϕ
2n·lnΓ2l+1
4n+ab −ϕ
2πn
with |Re ϕ|π/2, and where the right-hand side should be regarded as a limit in
cases ϕ=0 and ϕ=±π/2(a=1).
Hint: For exercise (a), first, evaluate the auxiliary integral JRwith the help of
‰
0Im zπ
eαz
ch2z+sin2ϕdz. This will give
∞
ˆ
0
chαx
ch2x+sin2ϕdx =π(κα/2−κ−α/2)
4sinϕsin2ϕ+1·csc απ
2,|Reα|<2.
Then, use a similar procedure as described in no. 3. Analogously, for exercise (b),
show first that
∞
ˆ
0
chαx
ch2x−sin2ϕdx =πsin αϕ
sin 2ϕ·csc απ
2,|Re α|<2.
10
*Prove by the contour integration method that for any a0 and Reb>0
(a) ∞
ˆ
0
ln(x2+1)
sh2bx dx =2
bln b
π−π
2b−Ψb
π
58 I.V. Blagouchine
(b) ∞
ˆ
0
chbx ·ln(x2+1)
sh2bx dx =1
bΨ1
2+b
2π−Ψb
2π−π
b,
(c) ∞
ˆ
0
(1−chbx) ln(x 2+a2)
sh2bx dx =−2
bln 2π
b+Ψ1
2+ab
2π,
(d) ∞
ˆ
0
ln(x2+a2)
sh2bx +cos2ϕdx =2π
bsin2ϕln Γ1
2+ab+ϕ
π
Γ1
2+ab−ϕ
π+4ϕ
bsin 2ϕln π
b
⎧
⎪
⎨
⎪
⎩
|Re ϕ|<π
2,if a=1,
|Re ϕ|π
2,if a=1,
where in the last expression the right part must be considered as a limit for ϕ=0 and
ϕ=±π/2(a=1).
Nota bene: Formula (a), in the unsimplified form, can be found in Bierens de Haan
tables [62, Table 258-5] and in [28, no. 4.373-4]. Formulas (b)–(d) seem to be
new.
11*By using the contour integration method prove that if pis a rational part of b,
i.e. p=bm/n, where bis some positive parameter and numbers mand nare positive
integers, then for any a>0,
(a) ∞
ˆ
0
sh2px ·ln(x2+a2)
sh2bx dx =1
b1−πm
nctg πm
n·ln πn
b
+2mπ
bn
n−1
l=1
sin 2πml
n·lnΓl
n+ab
πn−1
bn
n−1
l=1
cos 2πml
n·Ψl
n+ab
πn
+1
bΨab
π−1
bnΨab
πn−1
blnn,
(b) ∞
ˆ
0
sh2px ·lnx
sh2bx dx =−πm
2bn ctg πm
n·ln πn
b
+mπ
bn
n−1
l=1
sin 2πml
n·lnΓl
n−1
2bln2b
πsin mπ
n−γ
2b
(c) ∞
ˆ
0
chpx ·ln(x2+1)
sh2bx dx =−π
b2−πm
bn ctg πm
2n·ln 2πn
b
Malmsten’s integrals and their evaluation by contour integration 59
+2mπ
bn
2n−1
l=1
sin πml
n·lnΓl
2n+b
2πn
−1
bn
2n−1
l=1
cos πml
n·Ψl
2n+b
2πn−1
bnΨb
2πn
where m<nin exercises (a) an (b), and m<2nin exercise (c). For continuous and
complex values of p, see no. 67.
Hint: For exercise (a), after having proved that
∞
ˆ
0
sh2αx
sh2xdx =1
2−απ
2ctgαπ, |Re α|<1,
use a similar procedure as described in exercise no. 3. When evaluating residues, do
not forget that function sh2mz
sh2nz has removable singularities at z=0, z=πi,z=
2πi and poles of the second order at z=iπl/n,l=1,2,3,...,n−1,n+1,...,
2n−1. At the final stage, split the sum over l=1,2,...,2n−1 in two sums of equal
lengths (over l=0,1,2,...,n−1 and over l=n, n +1,n +2,...,2n−1), then,
use duplication formulas for both Γ- and Ψ-functions, and finally, employ the Gauss’
multiplication theorem for the Ψ-function
Ψ(nz)=lnn+1
n
n−1
l=0
Ψz+l
n,z∈C,n∈N.
Result (b) is obtained from (a) by an appropriate limiting procedure. In its final form
it appears after this elegant simplification
n−1
l=1
cos 2πml
n·Ψl
n=nln2sin mπ
n+γ, m=1,2,...,n−1.(48)
Formula (c) may be got from an intermediate formula obtained when deriving for-
mula (a) and with the help of no. 10a.
12
*In precedent exercises we saw that the evaluation of certain logarithmic integrals
by the contour integration method may lead to the Γ-function of a complex argument.
By supposing that parameters α,βand ϑare real and a0, show that integrals
(a) ∞
ˆ
0
ln(x2+a2)
chx+ϑdx, (b) ∞
ˆ
0
ln(x2+a2)
ch2x+αdx,
(c) ∞
ˆ
0
ln(x2+a2)
sh2x+βdx
60 I.V. Blagouchine
will lead to the Γ-function of a real argument (if using the proposed contour integra-
tion method) only if
(a) −1<ϑ1,a=1,
−1ϑ1,a=1,(b) −1<α0,a= 1,
−1α0,a=1,
(c) 0<β1,a= 1,
0β1,a=1
respectively. Note that such integrals can be also calculated by using expansions in
geometric series (we employed such a method in exercises no. 18–21). By the way,
Malmsten et al. in [40] and [41] evaluated only such a kind of integral.
13
*By making use of the contour integration method show that for any a0, Reb>
0,
(a) ∞
ˆ
0
ln(x2+a2)
ch3bx dx =π
blnΓ3
4+ab
2π−lnΓ1
4+ab
2π+π
2bln 2π
b
+1
4πbΨ13
4+ab
2π−Ψ11
4+ab
2π,
(b) ∞
ˆ
0
chpx ·ln(x2+a2)
ch3bx dx =π(n2−m2)
2bn2sec mπ
2n·ln 2πn
b
−(n2−m2)π
bn2
2n−1
l=0
(−1)lcos (2l+1)mπ
2n·lnΓ2l+1
4n+ab
2πn
+m
bn2
2n−1
l=0
(−1)lsin (2l+1)mπ
2n·Ψ2l+1
4n+ab
2πn
−1
4bπn2
2n−1
l=0
(−1)lcos (2l+1)mπ
2n·Ψ12l+1
4n+ab
2πn
(c) ∞
ˆ
0
sh3px ·ln(x2+a2)
sh3bx dx =π
b·
3m2
n2−tg2mπ
2n
1−3tg2mπ
2n
·tg mπ
2n·ln 2πn
b
+π
b
2n−1
l=1
(−1)lsin3mπl
n·lnΓl
2n+ab
2πn
+3πm2
bn2
2n−1
l=1
(−1)lsin mπl
n·3cos2mπ l
n−1·lnΓl
2n+ab
2πn
Malmsten’s integrals and their evaluation by contour integration 61
+3m
bn2
2n−1
l=1
(−1)lsin2mπl
n·cos mπl
n·Ψl
2n+ab
2πn
+1
4bπn2
2n−1
l=1
(−1)lsin3mπl
n·Ψ1l
2n+ab
2πn
where p=bm/n, numbers mand nbeing positive integers such that m<3nin (b)
and m<nin (c).
Hint: For exercise (b): use the procedure described in the hint of exercise no. 3.For
the evaluation of the auxiliary integral JR, consider
‰
0Im zπ
eαz
ch3zdz, and then, show that
∞
ˆ
0
chαx
ch3xdx =π(1−α2)
4·sec απ
2,|Re α|<3.
As regards exercise (c), the procedure is very similar. The evaluation of the integral
JRmay be done with the help of
p.v.‰
0Im zπ
eaz
sh3zdz, |Rea|<3,which yields
∞
ˆ
0
sh3αx
sh3xdx =π
2·3α2−tg2απ
2
1−3tg2απ
2·tg απ
2,|Re α|<1.
14
*Prove by the contour integration method that for any a0 and Re b>0,
(a) ∞
ˆ
0
ln(x2+a2)
ch4bx dx =4
3bln π
b+Ψ1
2+ab
π+1
3π2bΨ21
2+ab
π,
(b) ∞
ˆ
0
(1−chbx)2ln(x2+a2)
sh4bx dx =2
3bln 2π
b+Ψ1
2+ab
2π
+1
6π2bΨ21
2+ab
2π,
(c) ∞
ˆ
0
chpx ·ln(x2+a2)
ch4bx dx =(4n2−m2)mπ
6bn3csc mπ
2n·ln 2πn
b
62 I.V. Blagouchine
−(4n2−m2)mπ
3bn3
2n−1
l=0
sin (2l+1)mπ
2n·lnΓ2l+1
4n+ab
2πn
+4n2−3m2
6bn3
2n−1
l=0
cos (2l+1)mπ
2n·Ψ2l+1
4n+ab
2πn
−m
4bπn3
2n−1
l=0
sin (2l+1)mπ
2n·Ψ12l+1
4n+ab
2πn
+1
24bπ2n3
2n−1
l=0
cos (2l+1)mπ
2n·Ψ22l+1
4n+ab
2πn
where p=bm/n, numbers mand nbeing positive integers such that m<4n.
Hint: For exercise (c): proceed similarly to exercise no. 13c. As regards the integral
JR, consider
‰
0Im zπ
eαz
ch4zdz in order to prove that
∞
ˆ
0
chαx
ch4xdx =απ(4−α2)
12 ·csc απ
2,|Re α|<4.
15
*Show by the contour integration technique that
∞
ˆ
0
ln(x2+a2)
ch5bx dx =3π
4blnΓ3
4+ab
2π−lnΓ1
4+ab
2π+3π
8bln 2π
b
+5
24πbΨ13
4+ab
2π−Ψ11
4+ab
2π
+1
192π3bΨ33
4+ab
2π−Ψ31
4+ab
2π,
provided that a0 and Reb>0.
16
*Prove that
∞
ˆ
0
ln(x2+a2)
ch6bx dx =16
15bln π
b+Ψ1
2+ab
π+1
3π2bΨ21
2+ab
π
+1
60π4bΨ41
2+ab
π,
provided that a0 and Reb>0.
Malmsten’s integrals and their evaluation by contour integration 63
17
*Show that for a0 and Reb>0
(a) ∞
ˆ
0
ln(x2+a2)
chnbx dx =πAn
blnΓ3
4+ab
2π−lnΓ1
4+ab
2π+1
2ln 2π
b
+1
b
1
2(n−1)
l=1
Bn,l
π2l−1Ψ2l−13
4+ab
2π−Ψ2l−11
4+ab
2π
for odd n, and
(b) ∞
ˆ
0
ln(x2+a2)
chnbx dx =An
bln π
b+Ψ1
2+ab
π+1
b
1
2n−1
l=1
Bn,l
π2lΨ2l1
2+ab
π
for even n,Anand Bn,l being some rational coefficients. After some effort, one can
obtain their numeric values for any positive integer n. Table 1gives these coeffi-
cients for nup to 20. Curiously enough, Anfor odd nare equal to the coefficients
in the Maclaurin expansion of 2(1−x)−1
2, while A−1
nfor even nare equal to the
coefficients in the Maclaurin expansion of 1
2(1−x)−3
2.
4.1.2 Further results obtained by a combination of various methods
18
*By using geometric series expansions and term-by-term integration, prove
(a) ∞
ˆ
0
xalnx
ebx −1dx =Γ(a+1)
ba+1Ψ(a+1)ζ(a +1)+ζ(a +1)−ζ(a +1)ln b,
Rea>0,
(b) ∞
ˆ
0
xalnx
ebx +1dx =(1−2−a)Γ(a+1)
ba+1Ψ(a+1)ζ(a +1)+ln2
2a−1ζ(a +1)
+ζ(a +1)−ζ(a +1)lnb,Re a>−1,a=0,
(c) ∞
ˆ
0
lnx
ebx +1dx =−(ln 2 +2lnb) ln2
2b,
(d) ∞
ˆ
0
ln2x
ebx +1dx =ln2
b−2γ1−γ2+1
3ln22+ln2 ln b+ln2b+π2
6,
64 I.V. Blagouchine
n AnBn,1Bn,2Bn,3Bn,4Bn,5Bn,6Bn,7Bn,8Bn,9
12– – – – – – – – –
311
4–– – – – – – –
53
45
24 1
192 –––– – – –
75
8259
1440 7
1152 1
23040 ––– – – –
935
64 3229
20160 47
7680 1
15360 1
5160960 –– – – –
11 63
128 117469
806400 17281
2903040 209
2764800 11
30965760 1
1857945600 –– – –
13 231
512 7156487
53222400 1997021
348364800 28067
348364800 871
1857945600 13
11147673600 1
980995276800 –– –
15 429
1024 2430898831
19372953600 1206053
218972160 230443
2786918400 8521
15606743040 11
6370099200 1
392398110720 1
714164561510400 ––
17 6435
16384 60997921
516612096 24615717239
4649508864000 15313957
183936614400 5599613
9364045824000 4097
1872809164800 493
117719433216000 17
4284987369062400 1
685597979049984000 –
19 12155
32768 141433003757
1264666411008 426540447313
83691159552000 249938765093
3012881743872000 391080857
618027024384000 574123
224737099776000 85177
14832648585216000 1843
257099242143744000 19
4113587874299904000 1
839171926357180416000
n AnBn,1Bn,2Bn,3Bn,4Bn,5Bn,6Bn,7Bn,8Bn,9
22– –– –– – – – –
44
31
3–– –– – – – –
616
15 1
31
60 ––– – – – –
832
35 14
45 1
45 1
2520 –– – – – –
10 256
315 164
567 13
540 1
1512 1
181440 –– – – –
12 512
693 3832
14175 139
5670 31
37800 1
90720 1
19958400 –– – –
14 2048
3003 1888
7425 148
6075 311
340200 1
64800 1
8553600 1
3113510400 –– –
16 4096
6435 17067584
70945875 33608
1403325 2473
2551500 67
3572100 41
224532000 1
1167566400 1
653837184000 ––
18 65536
109395 16229632
70945875 14956868
638512875 4679
4677750 2021
95256000 757
3143448000 23
15567552000 1
217945728000 1
177843714048000 –
20 131072
230945 263495168
1206079875 2919352
127702575 5839219
5746615875 21713
943034400 5473
18860688000 619
294226732800 17
1961511552000 1
53353114214400 1
60822550204416000
Tab le 1 Coefficients Anand Bn,l from exercise no. 17 (top table corresponds to exercise 17-a, bottom table to 17-b)
Malmsten’s integrals and their evaluation by contour integration 65
(e) ∞
ˆ
0
xalnx
sh bx dx =(2−2−a)Γ (a +1)
ba+1Ψ(a+1)ζ(a +1)+ln2
2a+1−1ζ(a +1)
+ζ(a +1)−ζ(a +1)lnb,Re a>0,
(f) ∞
ˆ
0
xalnx
chbx dx =Γ(a+1)
22aba+1!Ψ(a+1)−ln4b"ζa+1,1
4+ζa+1,1
4
−2a2a+2−1ln 2 ·ζ(a +1)−2a2a+1−1ζ(a +1)
+!Ψ(a+1)−ln4b"ζ(a +1),Re a>−1,a=0,
(g) ∞
ˆ
0
lnx
chbx dx =1
bγ1−γ11
4−3γln 2 −7
2ln22−πγ
2−π
2ln4b,
(h) ∞
ˆ
0
ln2x
chbx dx
=1
b−γ2+γ21
4+2γ1ln 2 +12γln22+9ln
32+2γγ
11
4
−2γγ
1+6γ2ln 2 +π3
12 +2γ11
4ln4b−2γ1ln b+6γln2lnb
+7ln
22lnb+πγ2
2+2πln22+π
2ln2b+πγ ln4b+2πln 2 lnb,
which hold for any b>0. Integrals containing higher powers of the logarithm in the
numerator27 may be easily evaluated by calculating the derivative with respect to a
of the corresponding right parts [e.g., results (d) and (h) were obtained in this way].
However, the derivation is usually long and quite tedious. As regards non-integers
powers of logarithm in the numerator, calculation of such integrals may be sometimes
carried out by other methods; for example, Malmsten [41] treated similar integrals by
a simple change of variable.
Nota bene: We have not found formulas (a), (b), (d)–(h) in Gradshteyn and Ryzhik’s
tables [28], neither in Prudnikov et al.’s [53] tables. However, these results are not
very complicated to derive, so, probably, they were presented elsewhere before.
Solution: all demonstrations are based on the summation of certain geometric series
which can be integrated term-by-term. For instance, in exercise d), which is perhaps
the most complicated, one should first represent ch−1xin the following form:
1
chx=2∞
n=0
(−1)ne−(2n+1)x ,Rex>0.
27Integer powers only.
66 I.V. Blagouchine
The latter series, being uniformly convergent, can be integrated term-by-term
∞
ˆ
0
xa
chbx dx =2∞
n=0
(−1)n∞
ˆ
0
xae−(2n+1)bx dx
=2Γ(a+1)
ba+1
∞
n=0
(−1)n
(2n+1)a+1=Γ(a+1)
2aba+1ηa+1,1
2,
Rea>−1, b>0. By recalling that η(s,v) can be reduced to ζ(s,v),see(4), and in
view of the fact that ζs, 1
2=(2s−1)ζ(s), the above formula reduces to
∞
ˆ
0
xa
chbx dx =Γ(a+1)
22aba+1ζa+1,1
4−2a2a+1−1ζ(a +1),
Rea>−1, b>0. Differentiating once/twice the latter expression with respect to a,
and then letting a→0, yields formulas (g)/(h) respectively. For more information
about the relationship between the Hurwitz ζ-function and the Stieltjes constants, see
exercises no. 63–64.
19
*By using results of the previous exercise, show that
(a) ∞
n=1
(−1)nlnbn
n=(γ −lnb) ln 2 −1
2ln22,b>0,
(b) ∞
n=1
(−1)n−1ln(2n−1)
2n−1=πlnΓ1
4−π
2ln 2 −3π
4lnπ−πγ
4.
Hint: For (a): consider no. 18-c and set b=1. Expand ch−1xas indicated in the hint
of no. 18, and then interchange the integration and summation. Proceed similarly for
series (b) and recall that integral no. 18-g is also the simplest Malsmten’s integral (1).
Nota bene: The logarithmic series from exercise a) is quite well known and can be
found in many sources, e.g. in [28, no. 4.325-1] or in [53, vol. I, p. 746, no. 5.5.1-
3]. Series (b) was derived by Malmsten in [41, unnumbered equation on the pp. 20
and 26] and also appears in [53, vol. I, p. 747, no. 5.5.1-6] in the form containing
Malmsten’s integral (1).
20
*Proceeding as above, prove that for any b>0
(a) ∞
ˆ
0
xalnx
chbx +cos ϕdx =2Γ(a+1)
ba+1sin ϕ
∞
n=1
(−1)n−1Ψ(a+1)−lnbn
na+1sinϕn,
a>−1,−π<ϕ<π,
Malmsten’s integrals and their evaluation by contour integration 67
(b) ∞
n=1
(−1)nlnbn ·sin ϕn
n=πlnΓ1
2+ϕ
2π+π
2lncos ϕ
2−π
2lnπ
+ϕ
2γ+ln 2π
b,−π<ϕ<π,
(c) ∞
n=1
lnbn ·sin ϕn
n=πlnΓϕ
2π+π
2ln sin ϕ
2−π
2lnπ
+ϕ−π
2γ+ln 2π
b,0<ϕ<2π,
(d) ∞
n=1
ln(2n−1)·sin[(2n−1)ϕ]
2n−1=π
2lnΓϕ
2π−lnΓ1
2+ϕ
2π
−π
4ln2π+γ−ln tg ϕ
2,0<ϕ<π,
(e) ∞
n=1
(−1)nln(2n−1)·sin[(2n−1)ϕ]
2n−1=1
4Φϕ−π
2−Φϕ+π
2
+γ
2lntgϕ
2+π
4−πtgϕ
42lnΓϕ
2π+1
4−2lnΓϕ
2π+3
4
+lntgϕ
2+π
4−ln2π,−π
2<ϕ< π
2,
where Φ(ϕ) is defined in exercise no. 21.
Hint: Formula (b) may be obtained by comparing result (a) with that obtained pre-
viously in exercise no. 2. Expansion (c) may be obtained from (b) by a shifting ϕ.
The difference between (c) and (b) with parameter b=1 yields (d). Analogously,
formula (e) may be deduced from no. 21-c.
Nota bene: Formulas (b) and (c), in a slightly different form, were previously derived
by Malmsten et al. [40, p. 62] and [41, p. 25, eqs. (64)–(65)]. They also appear in [53,
vol. I, p. 748] as no. 5.5.1-25 and 5.5.1-24 respectively; however, formula (b) appears
incorrectly in no. 5.5.1-25, and (c) in no. 5.5.1-24 contains an additional modulus
that must be removed. Once again, we regret that Prudnikov et al. [53] do not spec-
ify sources. As regards expansion (c), which actually represents the Fourier series
expansion for the logarithm of the Γ-function, this important result is attributed to
Ernst Eduard Kummer [39,p.86],[71, p. 250], [9, vol. I, p. 23], albeit he obtained
this expression only in 1847 [35, p. 4], i.e. 5 years later after the publication of the
Malmsten et al.’s dissertation [40], see Fig. 2.28 Moreover, taking into account that
28Strictly speaking, Kummer’s result [35, p. 4] is obtained from the Malmsten et al.’s one [40,formula
(74), p. 62] by putting a=π(2x−1).
68 I.V. Blagouchine
expansion no. 21-e was also derived by Malmsten et al. [40, p. 74], the authorship
of such a kind of series should without doubt be, attributed to Malmsten and not to
Kummer. As regards formulas (a), (d), and (e), they seem to be unpublished yet.
21*Analogous to the previous exercise, prove the following results related to sums
containing cosines and logarithms:
(a) ∞
n=1
(−1)nlnbn ·cos ϕn
n=(γ −lnb) ln2 cos ϕ
2+Φ(ϕ)
2
+πctgϕ
22lnΓ1
2+ϕ
2π+lncos ϕ
2+ϕ
πln2π−ln π
−π<ϕ<π
(b) ∞
n=1
lnbn ·cos ϕn
n=(γ −lnb) ln2sinϕ
2+Φ(ϕ −π)
2
+πctgϕ
22lnΓϕ
2π+ln sin ϕ
2+ϕ−π
πln2π−ln π
0<ϕ<2π,
(c) ∞
n=1
ln(2n−1)·cos[(2n−1)ϕ]
2n−1=γ
2lntg ϕ
2+1
4Φ(ϕ −π)−Φ(ϕ)
+πctgϕ
42lnΓϕ
2π−2lnΓ1
2+ϕ
2π+lntg ϕ
2−ln2π
0<ϕ<π,
(d) ∞
n=1
(−1)nln(2n−1)·cos[(2n−1)ϕ]
2n−1=π
4(γ +3lnπ+2ln2−ln cos ϕ)
−π
2lnΓ1
4+ϕ
2πΓ1
4−ϕ
2π,−π
2<ϕ< π
2
b>0, and where we denoted by Φ(ϕ) the following improper integral:
Φ(ϕ) ≡∞
ˆ
0
e−xlnx
chx+cos ϕdx =2∞
ˆ
1
lnln x
x(x2+2xcosϕ+1)dx, −π<ϕ<π.
Nota bene: Formula (d) was previously derived by Malmsten et al. [40, p. 74] and
[41, p. 39]29 and does not appear, to our knowledge, in other sources. As regards the
other sums, they do not appear in Gradshteyn and Ryzhik’s tables [28], neither in
29Malmsten originally wrote a/2 instead of ϕ.
Malmsten’s integrals and their evaluation by contour integration 69
Prudnikov et al.’s tables [53] nor in other sources that we could found. The reader
may also remark that the integral Φ(ϕ) may be trivially expressed in terms of the
polylogarithm’s derivative. Although much less trivially, but it can be also expressed
in terms of the second-order derivatives of the Hurwitz ζ-function:
Φ(ϕ) =2ln2π·ln2 cos ϕ
2−πctgϕ2lnΓ1
2+ϕ
2π+lncos ϕ
2
+ϕ
πln2π−ln π+ζ0,1
2+ϕ
2π+ζ0,1
2−ϕ
2π.(49)
This formula straightforwardly follows from the comparison of no. 21-a to no. 22-a.
22*Results in exercises no. 18–21 are obtained by means of geometric series. An-
other way to treat such problems consists is to use the Mittag–Leffler theorem and
similar expansions. By proceeding in this manner, prove
(a) ∞
n=1
(−1)nlnn·cos ϕn
n
=lnπ·ln cos ϕ
2−1
2ln22+γln 2 +∞
ˆ
0
ch(ϕx/π) −1
xsh xlnxdx
Υ(ϕ)
=(γ +ln2π) ·ln2 cos ϕ
2+1
2ζ0,1
2+ϕ
2π+ζ0,1
2−ϕ
2π
=−1
2ln22+γln 2 +(γ +ln2π)ln cos ϕ
2−2Γ11
2+Γ11
2+ϕ
2π
+Γ11
2−ϕ
2π,−π<ϕ<π,
(b) ∞
n=1
lnn·cos ϕn
n=lnπ·ln sin ϕ
2−1
2ln22+γln 2 +Υ(ϕ−π)
=(γ +ln2π) ·ln2sin ϕ
2+1
2ζ0,ϕ
2π+ζ0,1−ϕ
2π,
0<ϕ<2π,
(c) ∞
n=1
ln(2n−1)·cos[(2n−1)ϕ]
2n−1=1
2(γ +ln2π) ·ln tg ϕ
2
+1
4ζ0,ϕ
2π+ζ0,1−ϕ
2π−ζ0,1
2+ϕ
2π−ζ0,1
2−ϕ
2π,
0<ϕ<π,
70 I.V. Blagouchine
(d) ∞
n=0
(−1)nln(2n+1)·sin[(2n+1)ϕ]
2n+1
=1
2ln π
2·lntgπ
4−ϕ
2+1
2
∞
ˆ
0
sh(2ϕx/π)
xchxln xdx, −π
2<ϕ< π
2
where ζ(s , v) stands for the second derivative of the Hurwitz ζ-function with respect
to s, and Γ1(x) stands for the antiderivative of the first generalized Stieltjes constant
γ1(x). Note that formulas (a)–(c) actually represent Fourier series expansions for the
sum of two second-order derivative of the Hurwitz ζ-function at s=0 and have a
comparatively simple form. At the same time, they can be regarded as the reflection
formula for the second-order derivative of the Hurwitz ζ-function at s=0.
Hint: For (a): by using a theorem from [23, no. 27.09], one can show that the follow-
ing expansion holds (see also [23, no. 27.10.1, p. 265] and [29, no. 641, p. 73])
sh αz
sh z=−2π∞
n=1
(−1)nn·sin(απ n)
z2+π2n2,−1<α<1 (50)
and is valid in the entire complex plane except at points z=πin,n∈Z. Computing
the antiderivative of the above expansion with respect to α, and then bearing in mind
that
∞
ˆ
0
lnx
x2+π2n2dx =lnπn
2n, n>0,(51)
as well as using the series from no. 19-c, yields the first part of the formula (that
containing the integral Υ(ϕ)). The second part of the formula (that containing Hur-
witz ζ-functions) is derived as follows. First, as in no. 18, write sh−1xas a sum of a
geometric series. Then, apply term-by-term integration30 and utilize the integral def-
inition of the Hurwitz ζ-function, see e.g. [9, vol. I, p. 25, Eq. 1.10(3)], [5, p. 251,
§12.3]. Proceeding in this manner yields
∞
ˆ
0
xa(chbx −1)
sh xdx =Γ(a+1)
2a+1ζa+1,1
2+b
2+ζa+1,1
2−b
2
−22a+1−1ζ(a +1)(52)
30The series being uniformly convergent.
Malmsten’s integrals and their evaluation by contour integration 71
a>−2, |Re b|<1. Differentiating the latter expression with respect to a, and then,
using an appropriate limiting procedure, we obtain
∞
ˆ
0
chbx −1
xsh xlnxdx=1
2ζ0,1
2+b
2+ζ0,1
2−b
2+3
2ln22
+ln 2 ·lnπ+(γ +ln2)lncos πb
2
|Re b|<1. The second part of the formula in its final form is now straightforward.
In order to obtain the third variant of the formula (that containing two antiderivatives
of the first Stieltjes constants) consider again Υ(ϕ), which is also the antiderivative
of no. 63-a with respect to pat p=ϕ/π. The constant of integration is easily deter-
mined by putting ϕ=0, which yields for the latter the value of −2Γ1(1/2). In order
to get result (b), write ϕ−πinstead of ϕin (a). For (d): it is sufficient to ascertain
that
chαz
chz=π∞
n=0
(−1)n(2n+1)cos!(n +1
2)απ "
z2+(n +1
2)2π2,−1<α<1,
which holds in the whole complex plane except at z=n+1
2πi,n∈Z. If necessary,
results (b)–(d) may be also written in terms of the antiderivatives of the first Stieltjes
constants.
23
*By using a similar method, prove that for any a0 and −π<ϕ<π,wehave
(a) ∞
n=1
(−1)nln(n +a) ·[cosϕn −1]
n
=lnπ·ln cos ϕ
2+1
2
∞
ˆ
0
ch(ϕx/π) −1
xsh xlnx2+π2a2dx,
(b) ∞
n=0
(−1)nln(n +a) ·sin!(n +1
2)ϕ"
2n+1
=1
2lnπ·ln tgπ
4−ϕ
4+1
4
∞
ˆ
0
sh(2ϕx/π)
xchxlnx2+π2a−1
22dx.
Prove that if ϕis a rational part of π,i.e.ϕ≡πm/n where mis integer and nis
positive integer, then for any a∈C(except points specified below) and k=2,3,4,...
(c) ∞
l=1
sin ϕl
l+a=−1
2n
2n−1
l=1
sin ϕl ·Ψl+a
2n
72 I.V. Blagouchine
(d) ∞
l=1
sinϕl
(l +a)k=(−1)k
(2n)k(k −1)!
2n−1
l=1
sinϕl ·Ψk−1l+a
2n
(e) ∞
l=0
cosϕl
l+a=−1
2n
2n−1
l=0
cosϕl ·Ψl+a
2n
(f) ∞
l=0
cosϕl
(l +a)k=(−1)k
(2n)k(k −1)!
2n−1
l=0
cosϕl ·Ψk−1l+a
2n
where in (c)–(d) a=−1,−2,−3,..., and in (e)–(f) a=0,−1,−2,...
Nota bene: Trigonometric series have been the subject of numerous studies since the
18th century; it is therefore very difficult to know if formulas (c)–(f) represent some
new contribution or not. We, however, remark that no closed-form expressions for
these series are given in Prudnikov et al.’s tables [53], nor in Gradshteyn and Ryzhik’s
tables [28].31
Hint: For (a) and (b), demonstrations are similar to no. 22, except that integral (51)
is replaced with
∞
ˆ
0
ln(x2+a2)
x2+ε2dx =π
εln(a +ε), a > 0,ε>0,
see, e.g., the solution for [59, no. 22, p. 187]. Result (c) is obtained by applying ∂2
∂ϕ∂a
to (a). After the differentiation, if ϕis a rational part of π, then the integral in the right
part of (a) coincide with the derivative of Malmsten’s integral no. 3-(a) with respect
to a; this leads to the Ψ-function in the right part. Shifting ϕby πyields the series (c)
in its final form. Result (d) is obtained from (c) by calculating its (k −1)th derivative
with respect to a. In like manner, we obtain (e) from (b), but result (c) should be also
used. Finally, extensions to a∈Cfollow from the principle of analytic continuation.
24 From the results obtained in no. 22 and no. 19, prove that
(a) ζ0,1
2=−3
2ln22−ln πln 2,
(b) ∞
ˆ
0
ln(x2+1/4)·arctg(2x)
e2πx −1dx =3
4ln22+(lnπ−1)ln 2
2−1
2.
Nota bene: Mark Coffey [18] suggested that it may be possible to obtain integral (b)
by contour integration. Regrettably, he did not indicate how to circumvent the prob-
31Prudnikov et al.’ tables provides, however, several formulas for the series (c) and (e) when ais rational
[53, vol. I, § 5.4.3].
Malmsten’s integrals and their evaluation by contour integration 73
lem of branch points that have both logarithm and arctangent (we pointed out the
importance of this problem in Sect. 3.1).
Hint: For (a): put ϕ=0 in no. 22-a and compare to no. 19-a. Another way to prove
the same result is to recall that ζs, 1
2=(2s−1)ζ(s), and then, to use these well-
known results ζ(0)=−1
2and ζ(0)=−1
2ln2π. For (b): compute the second deriva-
tive of the Hermite representation for the Hurwitz ζ-function with respect to sat
s=0 and v=1
2,see(3).
25*By combining various methods, prove that if pis a rational part of b,i.e.p=
bm/n, where bis some parameter with positive real part, and numbers mand nare
positive integers such that m<n, then
(a) ∞
ˆ
0
sh px ·lnx
ebx −1dx =−π
2bctg mπ
n·ln 2πn
b+π
b
n−1
l=1
sin 2mπl
n·lnΓl
n
−n
2bmγ+ln bm
n,
(b) ∞
ˆ
0
sh px ·lnx
ebx +1dx =π
2bcsc mπ
n·ln 2πn
b
−π
b
n−1
l=0
sin (2l+1)mπ
n·lnΓ2l+1
2n
+n
2bmγ+ln bm
n.
Hint: From these two well-known (at the time) integrals
∞
ˆ
0
chax ·sin rx
sh πx dx =sh r
2(chr+cos a) and ∞
ˆ
0
e−ωx sin rx dx =r
r2+ω2,
Legendre [64, vol. II, p. 189] derived, by using elementary transformations the value
of the following integral32
∞
ˆ
0
sinrx
e2πx −1dx =1
4cth r
2−1
2r,|Re r|<2π, r =0.
32This integral was first evaluated by Poisson in 1813 in the famous work [51, pp. 214–220], see also [12,
p. 240]. The work [51] was later republished in several volumes of the Journal de l’École Polytechnique,
see namely [52].
74 I.V. Blagouchine
By the same line of reasoning, one can show that
sh ax
sh bx =−2shωx
e2bx −1+e−ωx,chax
chbx =2shωx
e2bx +1+e−ωx,
chax
sh bx =2chωx
e2bx −1+e−ωx,sh ax
chbx =−2chωx
e2bx +1+e−ωx,
where a≡b−ω. Accounting for these elementary formulas and using previously
obtained integrals in exercise no. 3, as well as bearing in mind that
∞
ˆ
0
e−ωx lnxdx=−γ+lnω
ω,Reω>0,
we obtain both integrals (a) and (b). The final formulas are obtained by further sim-
plification with the help of the duplication formula for the Γ-function and with the
help of these results known from elementary analysis:
n−1
l=1
l·sin (2l+1)mπ
n=−n
2csc mπ
n,
n−1
l=1
l·sin 2mπl
n=−n
2ctg mπ
n
for m=1,2,...,n−1.
Nota bene: The evaluation of
∞
ˆ
0
chpx ·ln x
ebx +1dx
requires the knowledge of the following integral
∞
ˆ
0
sh px ·lnx
chbx dx.
The latter may be expressed by means of the first generalized Stieltjes constants, see
exercise no. 65.
26 By employing results obtained in exercise no. 25, prove that
∞
ˆ
0
e−bx ·thbx ·ln xdx =−2π
blnΓ1
4+π
2bln 4π3
b+γ+lnb
b,Re b>0.
Nota bene: alternatively, the same result may be derived by the series expansion
method. By recalling that ch−1xis the sum of a geometric progression, one may
easily show that
thx=1+2∞
n=1
(−1)ne−2nx ,Re x>0.
Malmsten’s integrals and their evaluation by contour integration 75
This expansion and the use of the result obtained in no. 19-d yield the above formula.
27 Show that
(a) ∞
ˆ
0
e−xlnx
ex+1dx =∞
ˆ
1
lnln x
x2(1+x) dx =1
2ln22−γ,
(b) ∞
ˆ
0
(chx−1)ln x
e2x−1dx =1
2
∞
ˆ
0
e−x·thx
2·lnxdx
=1
2
∞
ˆ
1
(x −1)lnln x
x2(x +1)dx =1
2γ−ln22.
4.2 Logarithmic integrals of lnln-type in combination with polynomials
An appropriate change of variable applied to integrals treated in the preceding ex-
ercises allows one to evaluate many beautiful ln ln-integrals. Below we give several
examples of such integrals, but the given list is far from exhaustive.
28
*Prove that
1
ˆ
0
x2−3x+1
1+x2+x4lnln 1
xdx =∞
ˆ
1
x2−3x+1
1+x2+x4ln ln xdx =πln 2
3√3.
Hint: Use formulas (44) and (45).
29 By using formula (37) show that
(a)
1
ˆ
0
xlnln 1
x
1+x4dx =∞
ˆ
1
xlnln x
1+x4dx =π
8ln 2 +3lnπ−4lnΓ1
4,
(b)
1
ˆ
0
xn−1lnln 1
x
(1+xn)2dx =∞
ˆ
1
xn−1ln ln x
(1+xn)2dx =1
2n−γ+ln π
2n,
(c)
1
ˆ
0
xlnln 1
x
1+x2+x4dx =∞
ˆ
1
xln ln x
1+x2+x4dx
=π
12√36ln2−3ln3+8lnπ−12 ln Γ1
3,
(d)
1
ˆ
0
lnln 1
x
1+√2x+x2dx =∞
ˆ
1
lnln x
1+√2x+x2dx
76 I.V. Blagouchine
=π
4√25ln2π−2ln(2+√2)−8lnΓ3
8,
(e)
1
ˆ
0
lnln 1
x
1−√2x+x2dx =∞
ˆ
1
lnln x
1−√2x+x2dx
=π
4√27ln2π−2ln(2−√2)−8lnΓ1
8,
(f)
1
ˆ
0
xlnln 1
x
1+√2x2+x4dx =∞
ˆ
1
xlnln x
1+√2x2+x4dx
=π
8√24ln2−2ln(2+√2)+5lnπ−8lnΓ3
8,
(g)
1
ˆ
0
xlnln 1
x
1−√2x2+x4dx =∞
ˆ
1
xln ln x
1−√2x2+x4dx
=π
8√24ln2−2ln(2−√2)+7lnπ−8lnΓ1
8,
(h)
1
ˆ
0
lnln 1
x
1+2xcosϕ+x2dx =∞
ˆ
1
ln ln x
1+2xcosϕ+x2dx
=π
2sinϕln⎧
⎨
⎩
(2π)ϕ
πΓ1
2+ϕ
2π
Γ1
2−ϕ
2π⎫
⎬
⎭,|Reϕ|<π.
Show that the function Q(x) in each of these integrals satisfies the functional rela-
tionship Q(x−1)=x2Q(x ).
Nota bene: Many of the above integrals for bounds [0,1]were already treated by dif-
ferent authors. However, none of them noticed that they can be equally taken from 1
to ∞and that they all obey Q(x−1)=x2Q(x). In particular, result (a), in a slightly
different form, as well as result (b) were presented by Adamchik [2] (it should be
noted however that both integrals may be easily obtained from Malmsten’s integrals
(1) and (46), respectively, by means of a simple change of variable). Result (c) is a
particular case of Malmsten’s integral (17)forn=3 (see also exercise no. 32 below),
and it was also independently evaluated in [44]. Integral (h) for real ϕwas evaluated
by Malmsten [41, p. 24] and also appears in [61, Table 190-9], in [62, Table 147-
9] and in [28, no. 4.325-7] (see also exercise no. 2above). Integrals (d) and (e) are
particular cases of (h) with ϕ=π/4 and ϕ=3π/4 respectively. Integrals (f) and (g)
may be deduced from integrals (d) and (e) respectively by making a suitable change
of variable; integral (g) was also independently evaluated in [44] (however the au-
thors did not simplify their result). By the way, formula (h) may provide many other
Malmsten’s integrals and their evaluation by contour integration 77
useful results. For instance, the result given in the Proposition 7.5 [44] may be ob-
tained directly from h) by differentiating it with respect to ϕ; formulas obtained in
examples 7.8–7.10 [44] follows immediately from such a derivative.
30*In [67, p. 314] Vardi, after having considered three basic Malmsten’s formulas
(1) and (2a), (2b), supposed that the multiplicative inverse of the argument of the
Γ-function is the degree in which the poles of the integrand are the roots of unity.33
Prove that such an assumption is not generally true for the integrals of the form
1
ˆ
0
lnln 1
x
ax2+bx +cdx (53)
where a,b,c are arbitrary real coefficients. Determine exact conditions under which
this assumption is true.
Proof From condition (27), which implies that Q(x−1)=x2Q(x) [see Sect. 3.2], it
follows that the above integral may be expressed in terms of the Γ-function if a=c.
Consider the case b∈(−2a,+2a). As shown in no. 12-a, in such a case integral (53)
may be written by means of the Γ-function of a real argument. Since −1cosϕ
+1 for any 0 ϕ2π, the integral in question may be always written in the form
analogous to no. 29-h, namely:
1
ˆ
0
lnln 1
x
1−2xcosϕ+x2dx =∞
ˆ
1
lnln x
1−2xcosϕ+x2dx
=− π
2sinϕϕ−π
πln2π−ln π+ln sin ϕ
2+2lnΓϕ
2π
where ϕ∈(0,2π). Accordingly to Vardi’s statement, the zeros of the denomina-
tor x1,2must be the (2π/ϕ)th roots of 1, i.e., x
2π
ϕ
1,2=1. Computing the roots of
the quadratic polynomial in the denominator yields x1,2=e±iϕ. Hence x
2π
ϕ
1,2=
e±2πi =1, and thus, Vardi’s hypothesis is true. Consider now the case b/∈
(−2a,+2a). This case, in virtue of what was established in no. 12-a, leads to the
Γ-function of a complex argument (integrals no. 4,5,7-a, 8-a and 9-a are typical
33In fact, Vardi was not very clear in defining his idea of the relationship between the poles of the in-
tegrand and the argument of the Γ-function with the help of which Malmsten’s integrals are expressed.
The statement “in Eq. (2a) the number 3 plays the ‘key role’ and in Eq. (2b) 6 is the ‘magic number”’
[67, p. 313] may be also interpreted in the sense that the least possible integer in the denominator of the
argument of the Γ-function (and not the inverse multiplicative of the argument of the Γ-function) should
be equal to the degree in which the poles of the integrand are the roots of unity. However, the fallacy of
this statement is also evident from the proof given above.
78 I.V. Blagouchine
examples of such a case). More precisely, integral (53) reduces to34
1
ˆ
0
lnln 1
x
1+2xcht+x2dx =∞
ˆ
1
lnln x
1+2xcht+x2dx
=− πi
2shtit
πln2π−ln π+ln ch t
2+2lnΓ1
2+it
2π
with t∈(0,∞). By Vieta’s formulas, we easily find the roots of the quadratic poly-
nomial in the denominator: x1,2=−e∓t. Are these values the pth roots of 1? where
p≡1
1
2+it
2π=2π2
π2+t2−i2πt
π2+t2.
The straightforward verification shows that only xp
1=1, while xp
2= 1, and thus,
Vardi’s assumption is false. Now, consider the case b=2a. In this case, the quadratic
polynomial ax2+bx +ctakes the form a(x +1)2, and hence (53) reduces to Malm-
sten’s integral (46), which is given in terms of the Euler’s constant γand not of the
logarithm of the Γ-function. However, Vardi remarked that his assumption remains
applicable only if ax2+bx +cis irreducible, which is obviously not the case if
b=2a. Finally, when b=−2aintegral (53) does not converge.
As regards the case a=c, the general procedure for the evaluation of (53)isnot
yet well known, but in many cases such integrals have higher transcendence than
the Γ-function (see, e.g. the last paragraph in Sect. 3.2). Consequently, Vardi’s as-
sumption for integral (53) remains true only if coefficients a,b, c are chosen so that
a=cand −2a<b<+2a, where we may suppose, without loss of generality, that
a>0.
31
*With the help of (37) show that
1
ˆ
0
xα−1lnln 1
x
1+x2αdx =∞
ˆ
1
xα−1ln ln x
1+x2αdx =π
4αln 4π3
α−4lnΓ1
4,α>0.
Note that case α=1 corresponds to the simplest Malmsten’s integral (1); case α=2
gives the result obtained by Adamchik (see the previous exercise). Cases for other α
(not necessarily integer) seem to be new (at least, in this form). The above result may
be therefore regarded as a generalization of Malmsten’s integral (1).
32
*Show that if
1
ˆ
0
xδlnln 1
x
1±xα+xβdx =∞
ˆ
1
xδln ln x
1±xα+xβdx,
34To assure the convergence, we should take in the denominator 1+2xcht+x2rather than 1 −2xch t+
x2.
Malmsten’s integrals and their evaluation by contour integration 79
then, coefficients α,βand δshould be chosen so that β=2αand δ=α−1. Prove
then that
1
ˆ
0
xα−1lnln 1
x
1+xα+x2αdx =∞
ˆ
1
xα−1lnln x
1+xα+x2αdx =2π
3α√3ln 4π2
4
√27α2−3lnΓ1
3
1
ˆ
0
xα−1lnln 1
x
1−xα+x2αdx =∞
ˆ
1
xα−1lnln x
1−xα+x2αdx =4π
3α√3ln 4π2
4
√54α2−3lnΓ1
3
where α>0. Note that these integrals for α=1 give Malmsten’s integrals (2a),
(2b)[or(44) and (45)], respectively. Hence, the above formulas may be seen as a
generalization of (2a), (2b). Moreover, in the case α=2, these integrals become
Malmsten’s integrals (17) and (18) with n=3, respectively.
Hint: Establish the condition under which Q(x−1)=x2Q(x ). Then, make a change
of variable x=et/α.
33 Show that the function Q(x) of Malmsten’s integral (17) satisfies the functional
relationship Q(x−1)=x2Q(x ). Show that so do functions
xm−1−x−m−1
xn−x−nand xm−1+x−m−1
xn+x−n
where nand mare natural numbers.
Nota bene: Integrals containing such functions Q(x) were evaluated by Malmsten
in [41, pp. 7, 29].
34 Prove that for any positive α
(a)
1
ˆ
0
xαn
2−1lnln 1
x
1+xα+x2α+···+xnα dx =∞
ˆ
1
xαn
2−1ln ln x
1+xα+x2α+···+xnα dx
=2
αYn+1−π
2n+2tg π
2n+2·ln α
2,n=1,2,3,...,
(b)
1
ˆ
0
xαn
2−1lnln 1
x
1−xα+x2α−···+xnα dx =∞
ˆ
1
xαn
2−1ln ln x
1−xα+x2α−···+xnα dx
=2
αXn+1−π
2n+2sec π
2n+2·ln α
2,n=2,4,6,...,
where Ynand Xnare Malmsten’s integrals (17) and (18), respectively.
80 I.V. Blagouchine
Hint: For exercise (a), show first that
1
ˆ
0
xn−2
1+x2+x4+···+x2n−2dx =∞
ˆ
1
xn−2
1+x2+x4+···+x2n−2dx =π
2ntg π
2n
As regards exercise (b), follow the same line of reasoning.
35
*A family of logarithmic integrals In
In=
1
ˆ
0
xn−1lnln 1
x
(1+x2)ndx =∞
ˆ
1
xn−1lnln x
(1+x2)ndx =1
2n
∞
ˆ
0
lnx
chnxdx,
n=1,2,3,..., generates special mathematical constants in the following way:
I1=π
2ln 2 +3π
4lnπ−πln Γ1
4,
I2=−1
2ln 2 +1
4lnπ−γ
4,
I3=π
16 ln 2 +3π
32 lnπ−G
4π−π
8lnΓ1
4,
I4=⎧
⎪
⎪
⎨
⎪
⎪
⎩
−1
12 ln 2 +1
24 lnπ−γ
24 −7ζ(3)
48π2,[ζ-form],
−1
12 ln 2 +1
24 lnπ−γ
24 +1
96π2Ψ21
2,[Ψ-form],
I5=3π
256 ln 2 +9π
512 lnπ+π
768 −5G
96π−3π
128 lnΓ1
4−1
6144π3Ψ31
4,
I6=⎧
⎪
⎪
⎨
⎪
⎪
⎩
−1
60 ln 2 +1
120 lnπ−γ
120 −7ζ(3)
192π2−31ζ(5)
320π4,[ζ-form],
−1
60 ln 2 +1
120 lnπ−γ
120 +1
384π2Ψ21
2+1
7680π4Ψ41
2,[Ψ-form],
I7=5π
2048 ln 2 +15π
4096 lnπ+43π
92160
−259G
23040π−5π
1024 lnΓ1
4−7
147456π3Ψ31
4−1
2949120π5Ψ51
4,
Malmsten’s integrals and their evaluation by contour integration 81
I8=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
−1
280 ln 2 +1
560 lnπ−γ
560 −49ζ(3)
5760π2−31ζ(5)
960π4−127ζ(7)
1792π6,
[ζ-form],
−1
280 ln 2 +1
560 lnπ−γ
560 +7
11520π2Ψ21
2+1
23040π4Ψ41
2
+1
1290240π6Ψ61
2,[Ψ-form],
Prove the results above by the contour integration technique.
Nota bene: The particular case n=1 is the simplest Malmsten’s integral (1); case
n=2 is also known, see e.g., [62, Table 257-4], [28, no. 4.371-3], [44]. The result
for the case n=3 can be found in [44]. Integral I4was also treated in [44], but
the presented formula differs from the above ones and contains the derivative of the
Riemann ζ-function. As regards integrals with higher n, they seem never to have been
evaluated before in the literature.
36
*By using results of the previous exercise, show that following mathematical con-
stants may be defined by ln ln-integrals:
(a) G=π
2
1
ˆ
0
(x4−6x2+1)ln ln 1
x
(1+x2)3dx =π
2
∞
ˆ
1
(x4−6x2+1)ln ln x
(1+x2)3dx
(b) ζ(3)=8π2
7
1
ˆ
0
x(x4−4x2+1)lnln 1
x
(1+x2)4dx =8π2
7
∞
ˆ
1
x(x4−4x2+1)lnlnx
(1+x2)4dx
The first result was already presented by Adamchik [2], while the second one seems
to be new. Similar integral definitions for the Euler’s constant γand for ln Γ(1/4)
are straightforward from the previous exercise.
37
*By using results of exercise no. 4, show that
1
2sh1
+∞
ˆ
−∞
ln|x|
shx±sh 1 dx =∓ ∞
ˆ
0
lnx
sh2x−sh21dx
=∓4∞
ˆ
1
xlnln x
x4−2x2ch 2 +1dx =∓4
1
ˆ
0
xlnln 1
x
x4−2x2ch 2 +1dx
=±2π
sh 2 Imln Γi
2π−lnΓ1
2−i
2π+π2
2sh2 +2ln2π
sh 2 .
82 I.V. Blagouchine
38
*Prove that for any t>0
∞
ˆ
1
x(ln ln x−lnt)
x4−2x2ch2t+1dx =
1
ˆ
0
xlnln 1
x−lnt
x4−2x2ch2t+1dx
=− π
2sh2tImlnΓit
2π−lnΓ1
2−it
2π−π2
8sh2t+t
2sh2tln t
2π.
4.3 Arctangent integrals containing hyperbolic functions
Integrals of the arctangent function in combination with hyperbolic functions are not
presented at all in [28], and there are few of them in [53, vol. I, § 2.7]. For example,
integral no. 39-c is given in [53, vol. I] as no. 2.7.5-10, but the provided formula is
incorrect.35
39
*By using Cauchy’s residue theorem, prove that for any Rea>0 and Reb>0
(a) ∞
ˆ
0
arctgx
sh xdx =∞
ˆ
0
arctgln x
x2−1dx =2∞
ˆ
1
arctgln x
x2−1dx =2
1
ˆ
0
arctgln x
x2−1dx
=
1
ˆ
0
arctg(2arcthx)
xdx =∞
ˆ
0
arctg2arcthe−xdx =
π/2
ˆ
0
x
sh tg x·cos2xdx
=πlnΓ1
2π−lnΓ1
2+1
2π−1
2ln2π
=π2lnΓ1
2π−lnΓ1
π−lnπ√8+ln 2
(b)
1
ˆ
0
arctg arcth x
xdx =πlnΓ1
π−lnΓ1
2+1
π−1
2lnπ
(c)
1
ˆ
0
arcth arcth x
xdx =−i
i
ˆ
0
arctgarctg x
xdx
=−πiln Γ−i
π−lnΓ1
2−i
π−1
2lnπ−πi
4
35More precisely, the last term in the right-hand side of [53, vol. I, no. 2.7.5-10] is incorrect: the argument
of the square root should be multiplied by 2. Curiously, in the original Russian edition of [53], this integral
is neither correctly evaluated, but the error is not the same: the coefficient 2 in the argument of the logarithm
must be placed under the square root sign.
Malmsten’s integrals and their evaluation by contour integration 83
(d) ∞
ˆ
0
arctgx
sh πx dx =1
2ln π
2
(e) ∞
ˆ
0
arctgax
sh bx dx =π
blnΓb
2πa−ln Γ1
2+b
2πa−1
2ln 2πa
b
Nota bene: Although these formulas do not appear correctly in modern mathematical
literature, a particular case of one of them may be found in the Malmsten et al.’s
dissertation [40, p. 52, Eq. (63)] and it is correct. As usual, Malmsten et al. used the
series expansion technique in order to get the result.
40
*By using results of the previous exercise, prove that for any Re a>0 and Reb>
0
(a) this analog of Binet’s formula:
∞
ˆ
0
arctgax
ebx +1dx =−π
blnΓ1
2+b
2πa−1
2a1+ln 2πa
b+π
2bln2π,
(b) this analog of Legendre’s formula:
∞
ˆ
0
xdx
(ebx +1)(x2+a2)=∞
ˆ
0
xdx
(eax +1)(x2+b2)=1
2Ψ1
2+ab
2π−ln ab
2π
Hint: Make use of Binet’s formula for the logarithm of the Γ-function
lnΓ(z)=z−1
2lnz−z+1
2ln2π+2∞
ˆ
0
arctg(x/z)
e2πx −1dx, (54)
see [12, pp. 335–336] and [71, pp. 250–251], [9, vol. I, p. 22, Eq. 1.9(9)]. It is inter-
esting that Legendre was very close to this expression and Binet also remarked this.
On p. 190 [64, vol. II], we find this expression
∞
ˆ
0
xdx
(e2πx −1)(m2+x2)=−1
4m+1
2lnm−1
2Ψ(m), (55)
which is exactly the derivative of the above Binet’s formula with respect to zat z=m.
In order to arrive at Binet’s formula (54), it was just sufficient to integrate (55) over m
and to find the constant of integration, but Legendre left this honour to Binet. It seems
also almost incredible that it took 23 years to get formula (54) from (55). Finally, a
more general version of Binet’s formula containing ebx −1 in the denominator of the
integrand appears also in Prudnikov et al.’s tables [53, vol. I] as no. 2.7.5-6 and is
correct.
84 I.V. Blagouchine
By the way, Lerch in [37, p. 19, Eq. (30)] (written in Czech) gives a more general
case of formula (a)
∞
ˆ
0
e2πx cosϕ−1
e4πx −2e2πx cosϕ+1arctg u
xdx =−1
4lnΓu−ϕ
2π+lnΓu+ϕ
2π
+lnu−ϕ
2π+2u(1−lnu) +ln sin ϕ
2−lnπ
in which parameters uand ϕare such that 0 <u1 and 0 <ϕ<2πu. This formula
reduces to exercise a) when setting u=1 and ϕ=π/2.
41*By using the contour integration method, prove that if pis a rational part of b,
i.e. p=bm/n, where a0, Reb>0 and numbers mand nare positive integers
such that m<n, then
(a) ∞
ˆ
0
sh px ·arctgax
chbx dx
=π
b
2n−1
l=0
(−1)lsin (2l+1)mπ
2n·lnΓ2l+1
4n+b
2πan
(b) ∞
ˆ
0
chpx ·arctg ax
sh bx dx =π
b
2n
l=1
(−1)lcos mπl
n·lnΓl
2n+b
2πan
+π
2bln 2πan
b
(c) ∞
ˆ
0
sh px ·arctgax
chbx +cos ϕdx =− π
bsin ϕ
×
n−1
l=0sin (2l+1)mπ +mϕ
n·lnΓ2l+1
2n+b+aϕ
2πan
−sin (2l+1)mπ −mϕ
n·lnΓ2l+1
2n+b−aϕ
2πan
(d) ∞
ˆ
0
sh px ·arctgax
chbx +1dx
=−2πm
bn
n−1
l=0
cos (2l+1)mπ
n·lnΓ2l+1
2n+b
2πan
Malmsten’s integrals and their evaluation by contour integration 85
−1
bn
n−1
l=0
sin (2l+1)mπ
n·Ψ2l+1
2n+b
2πan
where in (c) |Re ϕ|<π,ϕ=0; see (d) for ϕ=0.
Nota bene: A particular case of formula (a) for b=πwas already derived by Malm-
sten et al. [40, p. 70, Eq. (83)]. He separately treated cases (m +n) odd and (m +n)
even and simplified the result in both cases (simplification is not the same, see e.g. ex-
ercise no. 48 where we also treat separately these two cases and performed such a
simplification). Other formulas obtained above seem to be never released before.36
42
*By letting a→∞in the previous exercise, prove formula (23).
43
*Prove by the contour integration method the following formulas:
(a) p.v.
+∞
ˆ
−∞
arctgx
sh x±sh tdx =π
2chtln1+t2−2ln2π
+2π
chtReln Γ1
2π+it
2π−lnΓ1
2+1
2π−it
2π,
(b) p.v.
+∞
ˆ
−∞
arctgax
sh bx ±sh bt dx =π
2bchbt ln1+a2t2−2ln 2πa
b
+2π
bchbt Reln Γb
2πa +ibt
2π−lnΓ1
2+b
2πa −ibt
2π,
(c) +∞
ˆ
−∞
arctgax ±arctg at
sh bx ±sh bt dx =π
2bchbt ln1+a2t2−2ln 2πa
b
+2tarctgat
chbt +2π
bchbt Reln Γb
2πa +ibt
2π−lnΓ1
2+b
2πa −ibt
2π
where a>0, b>0, t>0.
44
*Show that
(a) ∞
ˆ
0
1−xcthx
sh xarctg2x
πdx =π
21−π
2,
(b) ∞
ˆ
0
(1−xcthx)arctgax
sh xdx =1
2aΨ1
2πa−Ψ1
2+1
2πa+πa,
36However, it seems fair to remark that an integral quite similar to (b) was also evaluated by Malmsten
et al. [40, p. 55, Eq. (67)].
86 I.V. Blagouchine
where Rea>0.
Hint: Use results of no. 39.
45
*Prove the following results:
(a) ∞
ˆ
0
sh x
ch2xarctgxdx =∞
ˆ
0
(x2−1)arctg ln x
(x2+1)2dx
=2∞
ˆ
1
(x2−1)arctg ln x
(x2+1)2dx =2
1
ˆ
0
(x2−1)arctg ln x
(x2+1)2dx
=∞
ˆ
1
arctgarcch x
x2dx =
e
ˆ
1
arctgarcch 1
lnxdx
x
=
1
ˆ
0
arctgarcch 1
xdx =
tg1
ˆ
0
arctgarcch 1
arctgxdx
1+x2
=
th1
ˆ
0
arctgarcch 1
arcthxdx
1−x2=
π/2
ˆ
0
x·th tg x
chtg x·cos2xdx
=1
2Ψ3
4+1
2π−Ψ1
4+1
2π,
(b) ∞
ˆ
0
sh x
ch2xarctg2x
πdx =ln2,
(c) ∞
ˆ
0
sh bx
ch2bx arctgax dx =1
2bΨ3
4+b
2πa−Ψ1
4+b
2πa,
where a>0, Reb>0.
46
*Show that for any a0 and Reb>0
(a) ∞
ˆ
0
sh px ·arctgax
sh2bx dx =πm
bn lnΓb
2πan−1
2ln 2πan
b
+πm
bn
2n−1
l=1
cos mπl
n·lnΓl
2n+b
2πan
Malmsten’s integrals and their evaluation by contour integration 87
+1
2bn
2n−1
l=1
sin mπl
n·Ψl
2n+b
2πan
(b) ∞
ˆ
0
sh px ·arctgax
ch2bx dx =−πm
bn
2n−1
l=0
cos (2l+1)mπ
2n·lnΓ2l+1
4n+b
2πan
−1
2bn
2n−1
l=0
sin (2l+1)mπ
2n·Ψ2l+1
4n+b
2πan,
where p=bm/n and numbers mand nare positive integers such that m<2n.
47
*Show that for any a0 and Reb>0
(a) ∞
ˆ
0
shbx
ch3bx arctgax dx =1
2πbΨ11
2+b
πa,
(b) ∞
ˆ
0
(1−chbx) arctg ax
sh3bx dx =π
2blnΓ1
2+b
2πa−ln Γb
2πa
+π
4bln 2πa
b+1
4πbΨ11
2+b
2πa,
(c) ∞
ˆ
0
(1−chbx)2arctg ax
sh5bx dx =π
4blnΓb
2πa−ln Γ1
2+b
2πa
−π
8bln 2πa
b−1
6πbΨ11
2+b
2πa−1
96π3bΨ31
2+b
2πa.
48*(a) By the contour integration method, prove that if pis a rational part of b,
i.e. p=bm/n, where bis some positive parameter and numbers mand nare positive
integers such that m<3n, then
∞
ˆ
0
sh px ·arctgax
ch3bx dx
=π(n2−m2)
2bn2
2n−1
l=0
(−1)lsin (2l+1)mπ
2n·lnΓ2l+1
4n+b
2πan
+m
2bn2
2n−1
l=0
(−1)lcos (2l+1)mπ
2n·Ψ2l+1
4n+b
2πan
88 I.V. Blagouchine
+1
8bπn2
2n−1
l=0
(−1)lsin (2l+1)mπ
2n·Ψ12l+1
4n+b
2πan
or, if considering separately cases (m +n) odd and (m +n) even,
∞
ˆ
0
sh px ·arctgax
ch3bx dx
=1
b·
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
π(n2−m2)
2n2
n−1
l=0
(−1)l+1sin (2l+1)mπ
2n·lnΓ1
2+2l+1
4n+b
2πan
Γ2l+1
4n+b
2πan
+m
2n2
n−1
l=0
(−1)lcos (2l+1)mπ
2n·Ψ2l+1
4n+b
2πan
−Ψ1
2+2l+1
4n+b
2πan
+1
8πn2
n−1
l=0
(−1)lsin (2l+1)mπ
2n·Ψ12l+1
4n+b
2πan
−Ψ11
2+2l+1
4n+b
2πan,if m+nis odd;
π(n2−m2)
2n2
n−1
l=0
(−1)lsin (2l+1)mπ
2n·lnΓ2l+1
2n+b
πan
+m
n2
n−1
l=0
(−1)lcos (2l+1)mπ
2n·Ψ2l+1
2n+b
πan
+1
2πn2
n−1
l=0
(−1)lsin (2l+1)mπ
2n·Ψ12l+1
2n+b
πan,
if m+nis even.
(b) Following Malmsten’s idea of establishing relationships between the Γ-function
and its logarithmic derivative, see (23), prove this more general formula implying the
trigamma function:
n2−m2
4n2Ψ1
4+m
4n−Ψ1
4−m
4n−πtg mπ
2n
=n2−m2
n2
2n−1
l=0
(−1)lsin (2l+1)mπ
2n·lnΓ2l+1
4n
+m
πn2
2n−1
l=0
(−1)lcos (2l+1)mπ
2n·Ψ2l+1
4n
Malmsten’s integrals and their evaluation by contour integration 89
+1
4π2n2
2n−1
l=0
(−1)lsin (2l+1)mπ
2n·Ψ12l+1
4n+m
2n
which holds for any positive integers mand nsuch that m<3n. Show that the right
part, analogous to (24), may be transformed into elementary functions and thus does
not contain any additional information about the trigamma function.
Hint: First, put for simplicity b=1 and write the integral (a) in the following form:
∞
ˆ
0
shm
nx
ch3xarctg(ax ) dx
=⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
n
∞
ˆ
0
sh my
ch3ny arctg(any) d y, y ≡x
n,if m+nis odd,
n
2
∞
ˆ
0
sh(1
2my )
ch3(1
2ny ) arctg(1
2any)dy, y ≡2x
n,if m+nis even.
Then, by taking into account that both integrands are rational functions of eyand,
hence, are 2πi-periodic, apply formula (38) to each of these integrals.37 At the final
stage, put b=1, make a→∞, and then compare the answer with the integral
∞
ˆ
0
sh αx
ch3xdx =−α
2+1−α2
4Ψ1
4+α
4−Ψ1
4−α
4−πtg πα
2,
|Re α|<3. The latter result may be obtained by various methods. For instance, it
may be obtained by making use of the following integral
∞
ˆ
0
e−αx
shβxdx =2β−1·Bα+β
2,1−β,Re β<1,Re(α +β)> 0
which is, in turn, easily derived from the definition of the Euler’s B-functionbya
simple change of variable, see, e.g., exercise no. 31.14.3 from [23]. By the way, in
this book, we found several errors related to the latter kind of integrals. In exercise
no. 31.11.6, in the right part the coefficient 2β−1should be replaced by 2β−2.The
answer given in exercise no. 31.11.5 is wrong: the integral in the left part cannot be
expressed via the Euler’s B-function. Unfortunately, these errors were not corrected
in the second and the last edition of this book. Finally, as regards the simplification of
the right part of (b) in terms of elementary functions, it is sufficient to show that the
37Alternatively, it is also possible to consider only the upper integral and then study two cases: (m +n)
is odd and (m +n) is even. This method will lead to the same formula, albeit the calculation might seem
more tedious.
90 I.V. Blagouchine
pairwise summation of terms in each sum, i.e. the summation of the kind al+a2n−1−l,
where, for example, as regards the third sum
al≡(−1)lsin (2l+1)mπ
2n·Ψ12l+1
4n,
does not contain the Γ-function, nor polygamma functions.
49
*Analogously to the previous exercise, show that for any a0 and Re b>0
ˆ∞
0
sh px ·arctgax
ch4bx dx
=−(4n2−m2)πm
6bn3
2n−1
l=0
cos (2l+1)mπ
2n·lnΓ2l+1
4n+b
2πan
−4n2−3m2
12bn3
2n−1
l=0
sin (2l+1)mπ
2n·Ψ2l+1
4n+b
2πan
−m
8bπn3
2n−1
l=0
cos (2l+1)mπ
2n·Ψ12l+1
4n+b
2πan
−1
48bπ2n3
2n−1
l=0
sin (2l+1)mπ
2n·Ψ22l+1
4n+b
2πan
where p=bm/n and numbers mand nare positive integers such that m<4n.
50
*Prove that Catalan’s constant is the following limit:
G=1+lim
α→1α
ˆ
0
(1+6x2+x4)arctgx
x(1−x2)2dx +2arcthα−πα
1−α2.
Hint: Show first that
¨
R2
xsin(2xy/π)
(x2+π2)ch xsh ydx dy =8(1−G).
4.4 Integrals containing logarithm of the Γ-function or polygamma functions in
combination with hyperbolic functions
Such integrals are not presented at all in Gradshteyn and Ryzhik’s tables [28], nor
in the second and third volumes of Prudnikov et al.’s tables [53]. Moreover, to our
knowledge, all results given below are new.
Malmsten’s integrals and their evaluation by contour integration 91
51
*Prove that for any a0
+∞
ˆ
−∞
lnΓx
πi +a
chxdx =π(a −1)ln 2 −1
2lnπ+2lnΓ1
4+a
2.
Hint: Consider ‰
0Im zπ
lnΓz
πi +a
chzdz.
52
*Prove that
+∞
ˆ
−∞
Ψx
πi ±1
2
chxdx =−π(γ ±ln2)
Hint: Consider ‰
0Im zπ
Ψz
πi −1
2
chzdz.
53
*Prove that for any a>0
+∞
ˆ
−∞
Ψx
πi +a
chxdx =πln 2 +Ψ1
4+a
2
Hint: Consider ‰
0Im zπ
Ψz
πi +a
chzdz.
54*The formula in exercise no. 51 is valid for a0, while that in no. 53 is valid
only for a>0. A simple way to shown that it is not valid for a=0 is to put a=0into
no. 53, which yields for the corresponding integral π−1
2π−γ−2ln2. However,
numeric computation38 shows that this result is incorrect; the correct one is
p.v.
+∞
ˆ
−∞
Ψx
πi
chxdx =∞
ˆ
0
Ψx
πi +Ψ−x
πi
chxdx =π+1
2π−γ−2ln2.
Explain this paradox and prove the above result.
55
*Prove that for any a>0
+∞
ˆ
−∞
Ψnx
πi +a
chxdx =π
2nΨn1
4+a
2,n=1,2,3,...
38With the help of Maple 12.
92 I.V. Blagouchine
where Ψndenotes the nth polygamma function.
56
*Prove that
+∞
ˆ
−∞
Ψx
2πi −1
2
(x2+π2)ch xdx =+∞
ˆ
−∞
Ψ3
2−x
2πi
(x2+π2)ch xdx =12 −4γ−12 ln 2
π+γ−1.
Hint: Consider ‰
0Imz2π
Ψz
2πi −1
2
(z −πi)ch zdz.
57
*Prove that
(a) +∞
ˆ
−∞
Ψ±x
2πi
(x2+π2)ch xdx =+∞
ˆ
−∞
Ψ1±x
2πi
(x2+π2)ch xdx
=4−4γ−12ln 2
π+γ+2ln2,
(b) +∞
ˆ
−∞
Ψ1
2±x
2πi
(x2+π2)ch xdx =γ−4(γ +3ln2−2G)
π,
(c) +∞
ˆ
−∞
xΨ1
2±x
2πi
(x2+π2)ch xdx =±πi(2−3ln2),
(d) +∞
ˆ
−∞
xΨx
2πi −1
2
(x2+π2)ch xdx =i(4+3π−3πln 2 −8G),
(e) +∞
ˆ
−∞
xΨx
πi
(4x2+π2)chxdx =i
82π−π2−4ln2.
4.5 Exercises concerning the Γ-function at rational arguments and the Stieltjes
constants
58*Prove that the Γ-function of any rational argument may be always expressed
via a finite combination of Malmsten’s integrals Ta(0)from (10), a=mπ/n, and
elementary functions:
lnΓk
n=(n −2k)ln2πn
2n+1
2lnπ−ln sin πk
n+1
πn
n−1
m=1
Im,n ·sin 2πmk
n,
Malmsten’s integrals and their evaluation by contour integration 93
k=1,2,...,n−1, n∈N2, where, for brevity, we designated by Im,n the following
quantity:
Im,n ≡∞
ˆ
0
sh(m
nx)
sh xlnxdx=π
2tg πm
2n·lnn+n
1
ˆ
0
xm−1−x−m−1
xn−x−nlnln 1
xdx
=π
2tg πm
2n·lnn+n
∞
ˆ
1
xm−1−x−m−1
xn−x−nlnln xdx =π
2tg πm
2n·lnπ+Ta(0).
Hint: First, take the second equation from (13) and put: x=0, 2minstead of m
and 2ninstead of n. This trick makes the equation valid for any integer values of
mand nbecause 2m+2nis always even. Then, notice that the obtained expression
represents a kind of the discrete sine transform for finite-length sequences. Hence,
use the orthogonality property of sin(πml/n) over the discrete interval [1,n−1]
n−1
m=1
sinπml
nsinπmk
n=n
2δl,k,
l=1,2,...,n−1, k=1,2,...,n−1. Perform a simplification with the help of this
formula
n−1
m=1
tg πm
2n·sin πmk
n=(−1)k+1·(n −k).
which is valid for positive integrers kand nsuch that k2n−1. At the final stage,
rewrite the result for 2kinstead of k.
There is also another way to prove this formula. Remark first that
n−1
m=1
shmx
nsin2πmk
n=sin 2πk
n·sh x
2cos 2πk
n−ch x
n
k=1,2,...,n−1, x∈C. Then, apply the main formula from exercise no. 2.
59*Analogously to the previous exercise, prove that for any k=0,1,...,2n−1,
where nis a positive integer,
lnΓ2k+1
4n=(−1)k
4nln 2π2
n+ln2πn
2nn−2#k+1
2$−(−1)k
nlnΓ1
4
+1
2lnπ−ln sin π(2k+1)
4n−(−1)k
πn
n−1
m=0
Im,n ·cos (2k+1)mπ
2n,
94 I.V. Blagouchine
where
Im,n ≡∞
ˆ
0
chm
nx
chxln xdx=π
2sec πm
2n·lnn+n
∞
ˆ
1
xm−1+x−m−1
xn+x−nlnln xdx.
Hint: First, consider formula (b) obtained in exercise no. 3and put a=0 and b=1.
Then, make use of the following semi-orthogonality property:
n−1
m=0
cos (2l+1)mπ
2n·cos (2k+1)mπ
2n=n
2δl,k +n
2δl,2n−1−k+1
2,
l=0,1,...,2n−1, k=0,1,...,2n−1. At the final stage, use the reflection formula
for the Γ-function, the Gauss’ multiplication theorem and the fact that
n−1
m=0
sec πm
2n·cos (2k+1)mπ
2n=(−1)kn−2#k+1
2$,
for any k=0,1,...,2n−1.
60
*Similarly to the previous exercises, show that
lnΓk
n=(n −2k)ln2πn
2n+1
2lnπ−ln sin πk
n
+1
2π
n−1
m=1
γ+ln(m/n)
m·sin 2πmk
n+1
πn
n−1
m=1
Im,n ·sin 2πmk
n
k=1,2,...,n−1, n∈N2, where
Im,n ≡∞
ˆ
0
sh(m
nx)
ex−1lnxdx =n
2m−π
2ctg πm
nlnn+n
2
∞
ˆ
1
xm−1−x−m−1
xn−1ln ln xdx.
Hint: First, consider formula (a) obtained in exercise no. 25. Then, use the semi-
orthogonality of sin(2πml/n) over [1,n−1]
n−1
m=1
sin2πml
nsin2πmk
n=n
2δl,k −n
2δl,n−k
l=1,2,...,n−1, k=1,2,...,n−1. Finally, use the reflection formula for the
Γ-function and recall that
n−1
m=1
ctg πm
n·sin 2πmk
n=n−2k
where kand nare positive integers such that k<n.
Malmsten’s integrals and their evaluation by contour integration 95
61
*Prove that for k=1,2,...,n−1, n∈N2,
lnΓk
n=(n −2k)lnπn
2n+1
2lnπ−ln sin πk
n+2
πn
n−1
m=1
Im,n ·sin 2πmk
n,
where
Im,n ≡∞
ˆ
0
sh2m
nx
sh2xlnxdx.
Hint: Take formula (b) from no. 11 and set b=1. Multiply both sides by sin 2πmk
n
and then sum over m=1,2,...,n−1. Remarking that
n−1
m=1
m·sin 2πml
n·sin 2πmk
n=n2
4(δl,k −δl,n−k),
l=1,2,...,n−1, k=1,2,...,n−1, and that
n−1
m=1
m·ctg πm
n·sin 2πmk
n=n
2(n −2k),
for k=1,2,...,n−1, we obtain the above formula. Another way to prove the same
result is to show that
n−1
m=1
Im,n ·sin 2πmk
n=1
4sin 2πk
n
∞
ˆ
0
lnx
cos2πk
n−ch2x
n
dx,
k=1,2,...,n−1; x∈C, and then, to apply formula (37) to the latter integral.
62
*Let ϕl,n be some known function of discrete arguments land n, which is defined
at least for l=1,2,...,n−1 and n∈N2. It is obvious that any countable combina-
tion of functions ϕl,n with other known functions will be a finite and known quantity.
Such a quantity is, for example,
Φk,n =
n−1
l=1
ϕl,n sin πlk
n,k=1,2,...,n−1.
One may easily note that the above formula is actually the expansion of the func-
tion Φk,n terms of orthogonal sines with coefficients ϕl,n; in other words, Φk,n is
the discrete sine transform of the sequence ϕ1,n,ϕ
2,n,...,ϕ
n−1,n. By using inter-
mediate results of exercise no. 58, prove the following functional relationship on
Γ( 1
2n), Γ ( 2
2n), Γ ( 3
2n),...,Γ(1
2−1
2n):
96 I.V. Blagouchine
n−1
k=1
(−1)kΦk,n lnΓk
2n=1
2π
n−1
m=1
Im,n ϕm,n −1
2
n−1
l=1
ϕl,n σl,n
−ln(2π2n)
4
n−1
l=1
ϕl,n tg πl
2n+lnπ
4
n−1
l=1
(−1)l+nϕl,n tg πl
2n,
where
σl,n ≡
n−1
k=1
(−1)ksin πlk
nln sin πk
2n,
the integral Il,n was defined previously in no. 58, and the above formula is valid for
any n∈N2.
(b) Suppose that the discrete function Υk,n may be represented by the finite Fourier
series39
Υk,n =β0,n +
n−1
l=1αl,n sin 2πlk
n+βl,n cos 2πlk
n,k=1,2,...,n−1,(56)
where coefficients αl,n and βl,n are known and finite (alternatively, Υk,n can be re-
garded as a discrete Fourier transform). Prove then that the following functional rela-
tionship for the logarithm of the Γ-function holds:
n−1
k=1
Υk,n lnΓk
n=1
π
n−1
m=1
I2m−n,nαm,n +ln πn
2
n−1
l=1
αl,n ctg πl
n−1
2
n−1
l=1
αl,nςl,n
−lnπ
2
n−1
l=1
βl,n +(n −1)lnπ
2β0,n −1
2
n−1
l=0
βl,n ωl,n,
n∈N2, where we designated for brevity
ςl,n ≡
n−1
k=1
sin 2πlk
nlnsin πk
n,ω
l,n ≡
n−1
k=1
cos 2πlk
nln sin πk
n
and the integral Il,n was defined previously in no. 58. Note that coefficients βl,n
do not affect the “computability” of the integral I2m−n,n αm,n. Moreover, if all
coefficients αl,n =0, then the right part contains elementary functions only.
Nota bene: A close study of the above-derived formulas reveals several interesting
things. First of all, physically, all equations represent a kind of Parseval’s equations
of closure (i.e., the law of conservation of energy). Moreover, from exercise (b), we
can straightforwardly establish Parseval’s theorem containing the sum of ln2Γ(k/n)
in the left part, but the right part will contain products of integrals Im,n with different
indices whose evaluation seems to be quite difficult. Second, if one could find such a
39For more details of the finite Fourier series, see e.g. [30, Chap. 6].
Malmsten’s integrals and their evaluation by contour integration 97
discrete functions Φk,n or Υk,n that coefficients ϕm,n or αm,n being summed with the
integrand from Im,n or I2m−n,n respectively, gave a new and “computable” integral,
then, it should be possible to obtain a new functional relationship for the logarithm
of the Γ-function.
63*(a) Show that Malmsten’s integral from exercise no. 3-a for a=0 may be also
computed by means of the first generalized Stieltjes constant γ1(v)
∞
ˆ
0
sh px ·lnx
sh xdx =−1
2π(γ +ln2)tg πp
2+γ11
2−p
2−γ11
2+p
2,
and this result holds for continuous and complex values of psuch that |Re p|<1.
(b) By making use of the previous result, prove following functional relationships
for the derivatives of the Hurwitz ζ-functions and for the first generalized Stieltjes
constants:
(1)lim
s→1ζs, 1
2−m
2n−ζs, 1
2+m
2n=γ11
2+m
2n−γ11
2−m
2n
=2π
2n−1
l=1
(−1)lsin mπl
n·lnΓl
2n+π(γ +ln4πn)tg mπ
2n,
(2)lim
s→1ζs,1−m
n−ζs, m
n=γ1m
n−γ11−m
n
=2π
n−1
l=1
sin 2πml
n·lnΓl
n−π(γ +ln2πn)ctg mπ
n
where m=1,2,...,n−1, n∈N2, and where derivatives are taken with respect to
the first argument of ζ(s,v).
Nota bene: Formulas (b.1) and (b.2) are not new40 and may be directly obtained by
differentiating the functional equation for the Hurwitz ζ-function, see e.g. [5, p. 261,
§12.9], [45, Eq. (6)], and by taking into account that ζ(0,v) =lnΓ(v)+ζ(0)=
lnΓ(v)−1
2ln2π, see e.g. [9, vol. I, p. 26, Eq. 1.10(10)], [11, Eq. (3)] or [2, Eq. (3)].
Though these relationships do not appear to be completely novel, the derivation from
Malmsten’s results (dated 1842!) and from the Mittag–Leffler theorem seems to be
original.
Hint: On the one hand, it appears from exercise no. 22 that the integral Υ(ϕ)may be
calculated by means of the Hurwitz ζ-function. On the other hand, one may remark
that the derivative dΥ/dϕ at ϕ=πm/n, where mand nare positive integers such
that m<n, coincide with Malmsten’s integral from exercise no. 3. By taking into
account that
∂
∂xζs, f (x)=−f
x·2ζs+1,f(x)
+sζs+1,f(x)
,
40According to [2] formula (c) was first proved by Almkvist and Meurman in a private communication.
98 I.V. Blagouchine
where derivatives ζand ζ are taken with respect to s, one easily arrives at the first
part of formula (b.1) [the part without the Stieltjes constants]. Now, it is well known
that the Hurwitz ζ-function ζ(s,v) is a meromorphic function on the entire complex
s-plane and that its only pole is a simple pole at s=1 with residue 1. It can be,
therefore, expanded as a Laurent series in a neighborhood of s=1 in the following
way
ζ(s,v)=1
s−1−Ψ(v)+∞
n=1
(−1)n(s −1)n
n!γn(v), s =1.
The coefficients γn(v) appearing in the regular part of this expansion are called gen-
eralized Stieltjes constants.41 From this formula, it follows that in a neighborhood of
s=1ζ(s, v ) =−(s −1)−2−γ1(v) +O(s −1)and ζ(s, v ) =2(s −1)−3+γ2(v) +
O(s −1), and thus, the limit in (b.1) reduces to the difference between two Stieltjes
constants. This yields formula (b.1) in its final form, as well as explaining how for-
mula (a) was obtained. Now, formula (b.1) may be rewritten in a slightly different
way. Putting 2m−ninstead of m, and then, using the duplication formula for the
Γ-function, as well as bearing in mind that
2n−1
l=1
l·sin 2πml
n=−nctg mπ
n,m=1,2,...,n−1,
one arrives at formula (b.2).
64*(a) Show that Malmsten’s integral from exercise no. 3-b for a=0 may be also
evaluated by means of the first generalized Stieltjes constant γ1(v)
∞
ˆ
0
chpx ·ln x
chxdx =1
2γ11
2+p
2+γ11
2−p
2
−γ11
4+p
4−γ11
4−p
4
−1
2ln22+ln 2 ·Ψ1
2+p
2+π
2(γ +ln2)tg πp
2
−π
2(γ +2ln2)ctgπ
4−πp
4
where parameter pis assumed to be continuous and complex lying within the strip
|Re p|<1.
41This expansion for the Riemann ζ-function was first given by Stieltjes, and therefore, was written in
terms of constants γn≡γn(1), which were later called the Stieltjes constants. Constants γn(v) with ar-
bitrary vrepresent a more general case and occur when expanding the Hurwitz ζ-function instead of the
Riemann ζ-function; such constants are called generalized Stieltjes constants. For more information, see
[9, vol. I, p. 26, Eq. 1.10(9), and vol. III, §17.7, p. 189], [14], [66,p.16],[11,17,18,38,48].
Malmsten’s integrals and their evaluation by contour integration 99
(b) Prove that the following functional relationship between first derivatives of
the Hurwitz ζ-function, first generalized Stieltjes constants and the logarithm of the
Γ-function takes place:
lim
s→1ζs, m
n+ζs,1−m
n−ζs, m
2n−ζs, 1
2−m
2n
=lim
s→1ζs, m
n−ζs, m
2n
+lim
s→1ζs,1−m
n−ζs, 1
2−m
2n
=−γ1m
n+γ11−m
n−γ1m
2n−γ11
2−m
2n
=2π
n−1
l=0
sin (2l+1)mπ
n·lnΓ2l+1
2n−πcsc mπ
n·lnπn −ln22
+2ln2·Ψm
n−π(γ +ln2)ctg mπ
n−π(γ +2ln2)tg mπ
2n,(57)
m=1,2,...,n−1,n∈N2.
Nota bene: This formula, as we come to see later, permits one to evaluate the first
generalized Stieltjes constant at some rational arguments. First of all, the combination
of the above formula with that from exercise no. 63 yields another elegant result
2γ1m
n−γ1m
2n−γ11
2−m
2n
=2π
n−1
l=1
sin 2πml
n·lnΓl
n−2π
n−1
l=0
sin (2l+1)mπ
n·lnΓ2l+1
2n
+ln22−2ln2·Ψm
n+π(γ +ln4πn)tg mπ
2n,(58)
where m=1,2,...,n−1, n∈N2. Several important particular cases follow im-
mediately from this expression. Thus, putting m=1 and n=2 yields
γ11
2−γ11
4=−2πlnΓ1
4+3π
2lnπ+2πln 2 +5
2ln22+γ
2(π +2ln2).
Setting m=1 and n=3gives
γ11
3−γ11
6=−2π√3lnΓ1
3+ln22+(3ln3+2γ)ln2
+π
√35ln2+4lnπ−1
2ln 3 +γ,(59)
and so on. However, other particular cases look less beautiful.
100 I.V. Blagouchine
Now, we proceed with the evaluation of the generalized Stieltjes constants at some
rational arguments. Let us first calculate γ1(1/4). By chance, we have ζs, 1
2=
(2s−1)ζ(s). By expanding both sides in a neighborhood of s=1 and by equating
coefficients with same powers, we arrive at
γ11
2=−2γln2 −ln22+γ1=−1.353459680... (60)
where as usually γ1≡γ1(1)=−0.07281584548... Hence,
γ11
4=2πlnΓ1
4−3π
2lnπ−7
2ln22−(3γ+2π)ln 2 −γπ
2+γ1
=−5.518076350 ....
By the formula from exercise no. 63-b.2, we immediately get
γ13
4=−2πln Γ1
4+3π
2lnπ−7
2ln22−(3γ−2π)ln 2 +γπ
2+γ1
=−0.3912989024 ....
Our next step is the evaluation of γ1(1/3),γ1(2/3),γ1(1/6)and γ1(5/6).Byusing
elementary transformations, one can show that ζs, 1
3+ζs, 2
3=(3s−1)ζ(s), and
hence
γ11
3+γ12
3=−3γln3 −3
2ln23+2γ1.(61)
By no. 63-b.2, it follows immediately that
γ11
3=−3γ
2ln 3 −3
4ln23+π
4√3ln 3 −8ln2π−2γ+12 ln Γ1
3+γ1
=−3.259557515 ...,
γ12
3=−3γ
2ln 3 −3
4ln23−π
4√3ln 3 −8ln2π−2γ+12 ln Γ1
3+γ1
=−0.5989062842 ....
By substituting γ1(1/3)into (59), we obtain
γ11
6=−3γ
2ln 3 −3
4ln23−ln22−(3ln3+2γ)ln2 +3π√3lnΓ1
3
+π
2√33
2ln 3 −14ln2 −12ln π−3γ+γ1=−10.74258252 ....
In view of the fact that Γ(1/6)=31
22−1
3π−1
2Γ2(1/3), see e.g. [15, p. 31], the latter
formula may be also written as
Malmsten’s integrals and their evaluation by contour integration 101
γ11
6=−3γ
2ln 3 −3
4ln23−ln22−(3ln3+2γ)ln2 +3π√3
2lnΓ1
6
−π
2√33ln3+11 ln2 +15
2lnπ+3γ+γ1=−10.74258252 ....
Hence, by no. 63-b.2, we also have
γ15
6=−3γ
2ln 3 −3
4ln23−ln22−(3ln3+2γ)ln2 −3π√3
2lnΓ1
6
+π
2√33ln3+11 ln2 +15
2lnπ+3γ+γ1=−0.2461690038 ....
However, further evaluation of γ1(k/n) faces much more difficulties. To illustrate this
point, we first generalize equations (60) and (61). From elementary transformations,
it follows that
n−1
l=1
ζs, l
n=ns−1ζ(s), n =2,3,4,....
Writing the Laurent series about s=1 for both sides and equating coefficients with
same powers, one obtains42
n−1
l=1
γ1l
n=−nγ lnn−n
2ln2n+(n −1)γ1,n=2,3,4,.... (62)
Moreover, analogously it can be shown that
n−1
l=0
ζs,v +l
n=nsζ(s,nv), n=2,3,4,...
and hence
n−1
l=0
γ1v+l
n=nlnn·Ψ(nv)−n
2ln2n+nγ1(nv), n =2,3,4,.... (63)
The latter formula represents a kind of the multiplication theorem for the first Stieltjes
constants. It can be extended to the higher-order Stieltjes constants as follows:
n−1
l=0
γpv+l
n=(−1)pnlnn
p+1−Ψ(nv)
lnpn+n
p−1
r=0
(−1)rCr
pγp−r(nv) ·lnrn,
n=2,3,4, where, as usually, Cr
pdenotes the binomial coefficient Cr
p=p!
r!(p−r)!.
A particular case of this formula for v=1/n was already found by Mark Coffey [18,
p. 1830, Eq. (3.28)].
42This formula appears with an error in [18, p. 1836, Eq. (3.54)]: in the right part 1
2should be replaced by
1
2lnq.
102 I.V. Blagouchine
Now, another useful property of the generalized Stieltjes constants may be derived
from the recurrence formula for the Hurwitz ζ-function:
ζ(s,v +1)=ζ(s,v)−v−s.
Expanding both sides in the Laurent series about s=1, one can easily see that
γ1(v +1)=γ1(v) −ln v
v,(64)
and more generally
γp(v +1)=γp(v) −lnpv
v,p=1,2,3,.... (65)
This is the recurrence relationship for the pth generalized Stieltjes constant. Thus,
equations (63), (64), and no. 63-b.2 constitute the multiplication theorem, the re-
currence relationship and the reflection formula, respectively, for the first Stieltjes
constant43 and may be quite useful for the exact determination of some values of γ1.
Now, we try to evaluate the set of γ1(k/8),k=1,2,...,7. Taking into account
previously calculated values, there are 4 unknowns in this set. By making use of
various formulas derived before, we may construct the following system of equations
for these unknowns:
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
γ1(1/8)+γ1(3/8)=···see Eq. (58)form=1, n=4,
γ1(1/8)+γ1(3/8)+γ1(5/8)+γ1(7/8)=··· see Eq. (62)forn=8,
γ1(1/8)−γ1(7/8)=··· see no. 63-b.2 for m=1, n=8,
γ1(3/8)−γ1(5/8)=··· see no. 63-b.2 for m=3, n=8,
where all quantities in the right part may be expressed in terms of γ,γ1,theΓ-
function and elementary functions. However, this system cannot be solved because
the corresponding matrix is not of full rank:
rank⎛
⎜
⎜
⎝
11 0 0
11 1 1
10 0 −1
01−10
⎞
⎟
⎟
⎠=3.
The same problem of rank arises when trying to evaluate families γ1(k/5),k=
1,2,3,4, and many others.44 For instance, one may try to reuse the procedure of
determination of Γ(1/12)in terms of Γ(1/3)and Γ(1/4)described in [15, p. 31]
for γ1(1/12). An observation of the system of equations II and III [15, pp. 30–31]
shows that the rank of the corresponding matrix is 8, and thus, the system can be
43Note that these relationships are quite similar to those for the logarithm of the Γ-function.
44An attentive analysis shows that equations no. 63-b.2, (58)and(63)forn=2andv=m/(2n) [variable
ncorresponds to (58)] are linearly dependent.
Malmsten’s integrals and their evaluation by contour integration 103
solved in terms of Γ(1/3)and Γ(1/4),thevalueΓ(1/2)being known. An equiv-
alent system of equations for the Stieltjes constants has the rank equal to 7 (due to
the fact that the reflection formula for the first Stieltjes constant no. 63-b.2 slightly
differs from that for the Γ-function), and hence, it cannot be solved. Thus, for the
evaluation of other γ1(k/n), we need to get more independent equations than we cur-
rently possess. By the way, there is an interesting analogy between our results and
those obtained by Miller and Adamchik [45]. Those authors proposed a method for
the evaluation of ζ(−2k+1,p),k∈N, for some rational values of pin terms of di-
verse transcendental functions. In particular, they provided an explicit expression for
the case p=1
3, and concluded that the same method can be equally applied to cases
p=1
2,1
4,3
4,2
3,1
6and 5
6. At the same time, they also reported that the evaluation of
ζ(−2k+1,p),k∈N, for other rational values of pfaces some major problems.
Notwithstanding, taking into account previously derived results, we may conjec-
ture that any generalized Stieltjes constant of the form γ1(k/n), where kand nare
positive integers such that k<n, may be expressed by means of the Euler’s con-
stant γ, the first Stieltjes constant γ1, the logarithm of the –function at rational ar-
gument(s) and some relatively simple, perhaps elementary, function. This statement
may be written as follows:
γ1k
n=f(k,n,γ)+
n−1
l=1
αl(k, n) ·ln Γl
n+γ1,k=1,2,...,n−1,(66)
Note that a similar relationship exists for the 0th generalized Stieltjes constant, which
is simply γ0(k/n) =−Ψ(k/n), see e.g. [48, p. 153, Eq. (4.7)].
The results above suggest another interesting conjecture: the sum of the first Stielt-
jes constant at a rational argument with its reflected version may be expressed in
terms of some relatively simple function g(possibly elementary), the Euler’s con-
stant γand the first Stieltjes constant γ1. In other words
γ1k
n+γ11−k
n=g+2γ1,k=1,2,...,n−1,(67)
An interesting way to confirm or to refute this hypothesis could be to study the inte-
gral from exercise no. 66 for rational values of p.
Finally, we should say that above results seem to be novel though we do not know
for sure. For example, Mark Coffey [18] already remarked that constants γ1(1/3)and
γ1(2/3)may be separately written in terms of γ1[18, p. 1830, Collolary 1]; however,
he did not provide explicit expressions for them.45 In the same work, one may also
find several formulas for the differences of Stieltjes constants expressed in terms of
the second-order derivatives of the Hurwitz ζ-function, as well as several series and
integral representations for them.
45Moreover, the exact determination of γ1(1/3)and γ1(2/3)is based on a particular case of (3.28) at
k=1; such a case is given by (3.54) [18]. The latter equation, as we already noticed in footnote 42,
contains an error.
104 I.V. Blagouchine
Hint: The technique of proof is similar to that from the preceding exercise. First, by
using geometric series expansions, one can show that the following integral may be
expressed by means of alternating Hurwitz ζ-functions
∞
ˆ
0
xash bx
chxdx =Γ(a+1)
2a+1ηa+1,1
2−b
2−ηa+1,1
2+b
2,
a>−2, |Re b|<1, and hence, in virtue of (4), by means of ordinary Hurwitz
ζ-functions. Differentiating the above integral with respect to a, and then, letting
a→−1, yields
∞
ˆ
0
sh bx
xchxln xdx
=1
2ζ0,1
2+b
2−ζ0,1
2−b
2+ζ0,1
4−b
4−ζ0,1
4+b
4
+3b
2ln22+γbln 2 +2ln2·ln Γ1
2+b
2−(γ +ln2)ln cos πb
2
−ln 2 ·lnπ+(γ +2ln2)(1−b) ln 2 +2lnsinπ
4−πb
4,
where derivatives of the Hurwitz ζ-function are taken with respect to its first argu-
ment and where |Re b|<1. Now, the derivative of the latter integral with respect
to bcoincides, at b=m/n, with Malmsten’s integral from no. 3-b. Writing 2m−n
instead of m, and then simplifying the formula from no. 3-b for a=0, b=1 and
p=(2m−n)/n, produces the final result.
65*Prove that the following integral may be evaluated by means of generalized
Stieltjes constants
∞
ˆ
0
sh px ·lnx
chxdx =1
2π(γ +ln2)tg πp
2
−(γ +2ln2)Ψ1
4+p
4−Ψ1
4−p
4
+γ11
2−p
2−γ11
2+p
2−γ11
4−p
4+γ11
4+p
4,
where |Re p|<1. Note that if pis rational, then γ1(1/2−p/2)−γ1(1/2+p/2)may
be expressed in terms of the Γ-function (see no. 63-b.1). This integral may be also
important for the closed-form determination of the first generalized Stieltjes constant
(see the Nota bene of the next exercise).
Malmsten’s integrals and their evaluation by contour integration 105
Hint: See previous exercises. At the final stage, use the reflection formula for the
Ψ-function.
66
*Analogously to the previous exercise, prove that for |Re p|<1
∞
ˆ
0
(chpx −1)ln x
sh xdx =(γ +ln2)·Ψ1
2+p
2+ln 2 −π
2tg πp
2
+γ2+γ1−1
2γ11
2+p
2−1
2γ11
2−p
2.
Show then that this integral for p=1
2,1
3,2
3may be expressed in terms of elementary
functions and the Euler’s constant γ.
Nota bene: This integral plays an important role for the second conjecture (67) con-
cerning the first generalized Stieltjes constant (see exercise no. 64). The formula
given above is derived by making use of geometric series, which lead to the Hur-
witz ζ-function (see the hint below). It is, therefore, highly desirable to find another
method for the evaluation of this integral. One of the possibilities could be the appli-
cation of the Mittag–Leffler theorem to the integrand. Accordingly, we may expand
for any −1<p<1
chpz −1
sh z=2z∞
n=1
(−1)ncospπ n −1
z2+π2n2,z∈C,z= πni, n ∈Z
see e.g. [23, no. 27.10.2]. But the performance of term-by-term integration results in
a divergent series
∞
ˆ
0
(chpx −1)ln x
sh xdx =2∞
n=1
(−1)n(cospπ n −1)
∞
ˆ
0
xlnx
x2+π2n2dx
∞
=···.
One can also try to evaluate a similar integral
∞
ˆ
0
(chpx −1)xa−1
sh xdx =2∞
n=1
(−1)n(cospπ n −1)
∞
ˆ
0
xa
x2+π2n2dx
1
2πana−1sec 1
2πa
=πasec πa
2
∞
n=1
(−1)ncospπ n −1
n1−a.(68)
The equality holds for −1<p<1 and a∈(−1,+1), but it can be analytically
continued for a/∈(−1,+1). The integral is the analytic continuation of the sum
106 I.V. Blagouchine
for a1, while the sum analytically continues the integral for a−1. We obvi-
ously have to expect trouble with the right-hand part at a=±1,±3,±5,... because
of the secant. Since when a=−1,−3,−5,... the sum in the right-hand side con-
verges, these points are poles of the first order for the analytic continuation of inte-
gral (68). In contrast, for a=1,3,5,..., the integral on the left remains bounded,
and thus, these points are removable singularities for the right-hand side of (68). In
other words, formally (−1)n(cos pπn −1)na−1,n1, must vanish identically for
any odd positive a(exactly as η(1−a), the result which has been derived by Euler,
see e.g. [20, p. 85]). Thus, the partial differentiation of the right-hand side of (68)
with respect to aat a→1 leads to an A-summable divergent series (see [31, p. 7]).
When using divergent series methods, the major difficulty consists in the evaluation
of (−1)n(cospπ n −1)ln2n,n1, the series (−1)n(cospπn −1)ln n,n1,
being easily reducible to 1
2Ψ(1
2+1
2p) −1
4πtg 1
2πp +1
2γ+ln 2 (see e.g. [40,p.58,
Eq. (72)] or differentiate no. 20-b with respect to ϕ).
Another approach consists in the use either of polylogarithms, or of the Lerch
transcendent or of the hypergeometric function, but the price will be a high transcen-
dence. The search for more suitable analytic continuations of (68) remains, therefore,
relevant.
Hint: Differentiate formula (52) with respect to a, and then, let a→0. In order to
perform the latter limiting procedure, expand ζ-functions in the Laurent series (as
we did in no. 63). The final result is obtained by using the reflection formula for the
Ψ-function. As regards the values of the integral at p=1
2,1
3,2
3, use results obtained
in exercise no. 64.
67*In exercises no. 63–66, we saw that there is a connection between Malmsten’s
integrals of the first order and the first Stieltjes constants. Prove now that
∞
ˆ
0
sh2px ·ln x
sh2xdx
=1
2lnπ−ln sin πp +p!γ1(p) −γ1(1−p)"−(γ +ln 2)(1−πp ctg πp)
for |Re p|<1, and therefore, such a connection exists also between Malmsten’s in-
tegrals of the second order and the first Stieltjes constants.
Hint: From elementary analysis, it is well known that
1
y2−2y+1=∞
n=1
ny n−1,|y|<1.
By putting in the latter expansion y=e−2x, one can easily show that
∞
ˆ
0
xa·e−px
sh2xdx =4Γ(a+1)∞
n=1
n
(2n+p)a+1
Malmsten’s integrals and their evaluation by contour integration 107
=Γ(a+1)
2a−1ζa, p
2−p
2ζa+1,p
2,a>1,Re p>−2
and hence,
∞
ˆ
0
xachpx
sh2xdx =Γ(a+1)
2aζa, p
2−p
2ζa+1,p
2+ζa,−p
2
+p
2ζa+1,−p
2,a>1,|Re p|<2.
Furthermore, it can be analogously shown that
∞
ˆ
0
xash2px
sh2xdx =Γ(a+1)
2a+1ζ(a,p)−pζ (a +1,p)+ζ(a,−p)
+pζ(a +1,−p) −2ζ(a),a>−1,|Re p|<1.
Differentiating this integral with respect to a, and then letting a→0, as well as using
(65), produces the wanted result. By the way, by putting p=m/n, and by using
no. 63-b.2, we arrive at the result obtained in exercise no. 11-b.
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Ramanujan J (2017) 42:777–781
DOI 10.1007/s11139-015-9763-z
ERRATUM
Erratum and Addendum to: Rediscovery of Malmsten’s
integrals, their evaluation by contour integration
methods and some related results [Ramanujan J. (2014),
35:21–110]
Iaroslav V. Blagouchine1,2
Published online: 25 October 2016
© Springer Science+Business Media New York 2016
Erratum to: Ramanujan J (2014) 35:21–110
DOI 10.1007/s11139-013-9528-5
Addendum to Section 2.2
The historical analysis of functional Eqs. (20)–(22) on pp. 35–37 is far from exhaustive.
In order to give a larger vision of this subject, several complimentary remarks may be
needed.
First, on p. 37, lines 1–5, the text “By the way, the above reflection formula (21) for
L(s)was also obtained by Oscar Schlömilch; in 1849 he presented it as an exercise for
students [55], and then, in 1858, he published the proof [56]. Yet, it should be recalled
that an analog of formula (20) for the alternating…” may be replaced by the following
one: “By the way, between 1849 and 1858, the above reflection formula for L(s)was
also obtained by several other mathematicians, including Oscar Schlömilch [55, 56],
Gotthold Eisenstein [73, 84], and Thomas Clausen [75].1Yet, it should be noted that
formula (20) itself was rigorously proved by Kinkelin a year before Riemann [79,
p. 100], [78], and its analog for the alternating…”
1In 1849, Schlömilch presented the theorem as an exercise for students [55]. In 1858, Clausen [75]
published the proof to this exercise. The same year, Schlömilch published his own proof [56]. Eisenstein
did not publish the proof, but left some drafts dating back to 1849, see e.g. [73, 84].
The online version of the original article can be found under doi:10.1007/s11139-013-9528-5.
BIaroslav V. Blagouchine
iaroslav.blagouchine@univ-tln.fr; iaroslav.blagouchine@pdmi.ras.ru
1University of Toulon, La Garde, France
2Steklov Institute of Mathematics at St. Petersburg, St. Petersburg, Russia
123
778 I. V. Blagouchine
Second, on p. 37, in formula (22), the confusing part “n=1,2,3,...” should be
replaced by “n∈R”. In fact, by comparing values of η(1−n)to η(n)at positive
integers and by noticing that both of them contain the same Bernoulli numbers, Euler
deduced Eq. (22). After that, he carried out a number of complimentary verifications,
which suggested that Eq. (22) should hold not only for integer values of n, but also
for fractional and continuous values of n. Whence, he conjectured that (22) should
be true for any value of argument, including continuous values of n. In particular, on
p. 94 of [20], Euler wrote: “Par cette raison j’hazarderai la conjecture suivante, que
quelque soit l’exposant n, cette equation a toujours lieu :
1−2n−1+3n−1−4n−1+5n−1−6n−1+...
1−2−n+3−n−4−n+5−n−6−n+...
=−1·2·3···(n−1)(2n−1)
(2n−1−1)π cos πn
2.”
The latter is our Eq. (22) and is also equivalent to (20). By the way, Hardy’s exposition
of Euler’s achievements, which we cited in footnote 15, Ref. [31, pp. 23–26], is also
far from exhaustive. For instance, Hardy did not mention the fact that Euler have
conjectured that formula (22) remains true for any value of n. Moreover, Hardy says
that it was comparatively recently that it was observed, first by Cahen and then by
Landau, that the reflection formulas for L(s)and η(s)both stand in Euler’s paper
written in 1749. This is, however, not true. Thus, Malmsten in 1846 remarked [40,
p. 18] that (21) were obtained by Euler by induction, and Hardy cited this work of
Malmsten. Unfortunately, Hardy did not notice that Malmsten also quoted Euler.
Third, on p. 37, after the last sentence in the first paragraph ending by “…requires
the notion of analytic continuation.”, the following footnote may be added
“An alternative historical analysis of functional Eqs. (20)–(21) in the context
of contributions of various authors may be found in [85], [31, p. 23], [84],
[82, p. 4], [78, p. 193], [74, pp. 326–328], [83, p. 298], [73]. Note, however, that
Butzer et al.’s statement [74, p. 328] “Malmstén included the functional equation
without proof” is rather incorrect. Thus, André Weil [84, p. 8] points out that
“Malmstén included the proof in a long paper written in May 1846”. Moreover,
our investigations show that this proof was not only included in his paper [41]
written in 1846, but also was present in an earlier work [40] published in 1842.
By the way, Malmsten remarked that reflection formulas of such kind were first
announced by Euler in 1749, the fact which was not mentioned by Schlömilch
[55, 56], nor by Clausen [75], nor by Kinkelin [79], nor by Riemann [54].”
Addendum to Section 4.1.2, Exercise no. 18
Results of this exercise also permit to evaluate some very curious integrals containing
cos ln ln xand sinlnlnxin the numerator. Putting in the last unnumbered equation in
123
Erratum and addendum to “Rediscovery of Malmsten’s integrals”... 779
Exercise no. 18 on p. 66 a=iα,α∈R, and b=1, we have
∞
0
cos(α ln x)
chxdx=2
∞
1
cos(α ln ln x)
1+x2dx
=−αImΓ(iα)
22iαζ1+iα, 1
/4−2iα21+iα−1ζ(1+iα),
∞
0
sin(α ln x)
chxdx=2
∞
1
sin(α ln ln x)
1+x2dx
=αReΓ(iα)
22iαζ1+iα, 1
/4−2iα21+iα−1ζ(1+iα),
These integrals are, in some sense, complimentary to basic Malmsten’s integrals (1)–
(2), which were evaluated in Sect. 3.4 and 4.1.2, no. 18-g, and readily permit to evaluate
integrals
∞
0
lnnx
chxdx=2
∞
1
lnnln x
1+x2dx=2
1
0
lnnln 1
x
1+x2dx,n=1,2,3,...
in terms of Stieltjes constants (first two such expressions were given in Exercises
no. 18-g and 18-h). It is also interesting that right parts of both expressions contain
ζ(1+iα), which was found to be connected with the nontrivial zeros of the ζ-function.2
Addendum to Section 4.2, Exercise no. 29
In right parts of formulas (d)–(g), it may be more preferable to have ln(1+√2)rather
than ln(2±√2)
(d)
1
0
ln ln 1
x
1+√2x+x2dx=
∞
1
ln ln x
1+√2x+x2dx
=π
4√25lnπ+4ln2 −2ln(1+√2)−8lnΓ3
8,
(e)
1
0
ln ln 1
x
1−√2x+x2dx=
∞
1
ln ln x
1−√2x+x2dx
=π
4√27lnπ+6ln2 +2ln(1+√2)−8lnΓ1
8,
2Estimation of
ζ(1+iα)
was found to be connected with Reρ,whereρare the zeros of ζ(s)in the
critical strip 0 Res1, see e.g. [80, p. 128].
123
780 I. V. Blagouchine
(f)
1
0
xln ln 1
x
1+√2x2+x4dx=
∞
1
xln ln x
1+√2x2+x4dx
=π
8√25lnπ+3ln2 −2ln(1+√2)−8lnΓ3
8,
(g)
1
0
xln ln 1
x
1−√2x2+x4dx=
∞
1
xln ln x
1−√2x2+x4dx
=π
8√27lnπ+3ln2 +2ln(1+√2)−8lnΓ1
8.
Addendum to Section 4.5, Exercise no. 62-b
On p. 96, in Exercise no. 62-b, in the unnumbered formula after Eq. (56), in the first
line the last term
−1
2
n−1
l=1
αl,nςl,n
may be removed. Strictly speaking, the actual expression for Υk,nln Γk
nis cor-
rect. However, because of the symmetry, the function ςl,nidentically vanishes for any
integer l=1,2,...,n−1, and hence, so does the last term in the first line of this
formula.
Some minor corrections and additions
•p. 42, line 20: “has no branch points.” should read “has no branch points except
at poles of Γ(z).”
•p. 42, line 27: “points at all, which allows” should read “points at all in the right
half–plane, which allows”.
•p. 66, in Nota Bene of exercise no. 19: “derived by Malmsten in [41, unnumbered”
should read “derived by Malmsten in [40, p. 24, Eq. (37)], [41, unnumbered”.
•p. 68, first line: “no. 21-e” should read “no. 21-d”.
•p. 73, last line, “|Rer|<2π” should read “|Imr|<2π”.
•p. 82, line 7, “no. 39-c is given” should read “no. 39-e is given”.
•p. 83, exercise no. 40: formula given in exercise no. 40-b, as well as formula (55),
were also obtained by Nørlund in [81, p. 107].
•p. 97, footnote 40 “formula (c) was” should read “formula (b.2) was”.3
3It may also noted that in a later work [72, pp. 542–543], we showed that (b.1), which is a shifted version
of (b.2), was already known to Malmsten in 1846.
123
Erratum and addendum to “Rediscovery of Malmsten’s integrals”... 781
•p. 100, exercise no. 64: closed-form expressions equivalent to those we gave for
γ1(1
/2),γ1(1
/4),γ1(3
/4)and γ1(1
/3)were also obtained by Connon in [76, pp. 1,
50, 53, 54–55], [77, pp. 17–18].
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