A Simple Proof of the Riemann's Hypothesis

Abstract

We present a simple proof of the Riemann's Hypothesis (RH) where only undergraduate mathematics is needed.

Full text

A Simple Proof of the Riemann’s Hypothesis
Charaf ECH-CHATBI ∗
23 October 2023
Abstract
We present a simple proof of the Riemann’s Hypothesis (RH) where
only undergraduate mathematics is needed.
Keywords: Riemann Hypothesis; Zeta function; Prime Numbers;
Millennium Problems.
MSC2020 Classification: 11Mxx, 11-XX, 26-XX, 30-xx.
1 The Riemann Hypothesis
1.1 The importance of the Riemann Hypothesis
The prime number theorem gives us the average distribution of the primes.
The Riemann hypothesis tells us about the deviation from the average.
Formulated in Riemann’s 1859 paper[1], it asserts that all the ’non-trivial’
zeros of the zeta function are complex numbers with real part 1/2.
1.2 Riemann Zeta Function
For a complex number swhere ℜ(s)>1, the Zeta function is defined as
the sum of the following series:
ζ(s) =
+∞
X
n=1
1
ns(1)
In his 1859 paper[1], Riemann went further and extended the zeta function
ζ(s), by analytical continuation, to an absolutely convergent function in
the half plane ℜ(s)>0, minus a simple pole at s = 1:
ζ(s) = s
s−1−sZ+∞
1
{x}
xs+1 dx (2)
Where {x}=x−[x] is the fractional part and [x] is the integer part of x.
Riemann also obtained the analytic continuation of the zeta function to
the whole complex plane.
∗Email: charaf.chatbi@gmail.com. The opinions of this article are those of the author and
do not reflect in any way the views or business of his employer.
1
arXiv:submit/5224804 [math.GM] 11 Nov 2023

Riemann[1] has shown that Zeta has a functional equation1
ζ(s) = 2sπs−1sin πs
2Γ(1 −s)ζ(1 −s) (4)
Where Γ(s) is the Gamma function. Using the above functional equa-
tion, Riemann has shown that the non-trivial zeros of ζare located sym-
metrically with respect to the line ℜ(s)=1/2, inside the critical strip
0<ℜ(s)<1. Riemann has conjectured that all the non trivial-zeros are
located on the critical line ℜ(s) = 1/2. In 1921, Hardy & Littlewood[2,3,
6] showed that there are infinitely many zeros on the critical line. In 1896,
Hadamard and De la Vall´ee Poussin[2,3] independently proved that ζ(s)
has no zeros of the form s= 1 + it for t∈R. Some of the known results[2,
3] of ζ(s) are as follows:
•ζ(s) has no zero for ℜ(s)>1.
•ζ(s) has no zero of the form s= 1 + iτ. i.e. ζ(1 + iτ )= 0, ∀τ.
•ζ(s) has a simple pole at s= 1 with residue 1.
•ζ(s) has all the trivial zeros at the negative even integers s=−2k,
k∈N∗.
•The non-trivial zeros are inside the critical strip: i.e. 0 <ℜ(s)<1.
•If ζ(s) = 0, then 1 −s, ¯sand 1 −¯sare also zeros of ζ: i.e. ζ(s) =
ζ(1 −s) = ζ(¯s) = ζ(1 −¯s) = 0.
Therefore, to prove the “Riemann Hypothesis” (RH), it is sufficient to
prove that ζhas no zero on the right hand side 1/2<ℜ(s)<1 of the
critical strip.
1.3 Proof of the Riemann Hypothesis
Let us take a complex number ssuch that s=σ+iτ. Unless we explic-
itly mention otherwise, let us suppose that 0 < σ < 1, τ > 0 and ζ(s) = 0.
We have from the Riemann’s integral (2) above:
ζ(s) = s
s−1−sZ+∞
1
{x}
xs+1 dx (5)
We have s= 1, s= 0 and ζ(s) = 0, therefore:
1
s−1=Z+∞
1
{x}
xs+1 dx (6)
Or in other terms:
1
σ+iτ−1=Z+∞
1
{x}
xσ+iτ+1dx (7)
1This is slightly different from the functional equation presented in Riemann’s paper[1].
This is a variation that is found everywhere in the litterature[2,3,4]. Another variant using
the cos:
ζ(1 −s)=21−sπ−scos πs
2Γ(s)ζ(s) (3)
2

Let us denote the following functions:
ϵ(x) = {x}(8)
ϕ(x) = {x}1− {x}(9)
Ψ(x) = Zx
1
dx ϵ(x) (10)
K(σ, τ ) = τ
(1 −σ)2+τ2(11)
To continue, we will prove the following lemmas.
Lemma 1.1. Let us consider two variables σand τsuch that σ > 0and
τ > 0. Let us define two integrals I(a, σ, τ )and J(a, σ, τ )as follows:
I(a, σ, τ ) = Za
1
sin(τln(x))
xσdx (12)
J(a, σ, τ ) = Za
1
cos(τln(x))
xσdx (13)
Therefore
I(a, σ, τ ) = K(σ, τ )1−cos(τln(a))
aσ−1−(σ−1)
τ
sin(τln(a))
aσ−1(14)
J(a, σ, τ ) = K(σ, τ )(σ−1)
τ−(σ−1)
τ
cos(τln(a))
aσ−1+sin(τln(a))
aσ−1(15)
Proof. Let us consider two variables σand τsuch that σ > 0 and τ > 0.
Let us take a > 1.
I(a, σ, τ ) = Za
1
sin(τln(x))
xσdx (16)
=Zln(a)
0
sin(τx)e(1−σ)xdx (17)
=K(σ, τ )1−cos(τln(a))
aσ−1−(σ−1)
τ
sin(τln(a))
aσ−1(18)
And the same for J(a, σ, τ ) for a > 0:
J(a, σ, τ ) = Za
1
cos(τln(x))
xσdx (19)
=Zln(a)
0
cos(τx)e(1−σ)xdx (20)
=K(σ, τ )(σ−1)
τ−(σ−1)
τ
cos(τln(a))
aσ−1+sin(τln(a))
aσ−1(21)
Lemma 1.2. The function ϵ(x)is piecewise continuous on [0,+∞)and
its primitive function Ψ(x)is defined as follows:
Ψ(x) = 1
2x−1−ϕ(x)(22)
3

Let us consider two variables σand τsuch that 0< σ < 1and τ > 0such
that s=σ+iτis a zeta zero. Therefore:
Z+∞
1
dx Ψ(x)
x2+s=1
(s−1)(1 + s)(23)
And
Z+∞
1
dx ϕ(x)
x2+s=1
s(1 −s)(24)
Proof. We will need the function ϕas you will see later that we need a
continuous function instead of a piecewise one like the function ϵ.
Let us take x > 1 a real number. Let us denote nx=⌊x⌋be the integer
part of x. We have nx=x− {x}. Therefore, we can write the following:
Ψ(x) = Zx
1
ϵ(t)dt (25)
=
nx−1
X
n=1 Zn+1
n{t}dt +Zx
nx{t}dt (26)
=
nx−1
X
n=1 Zn+1
n
(t−n)dt +Zx
nx
(t−nx)dt (27)
=
nx−1
X
n=1
1
2+1
2(x−nx)2(28)
=1
2nx−1 + {x}2(29)
=1
2x−1− {x}+{x}2(30)
This prove the equation (23). ■
Let us prove the second point of the lemma. Let us define the integral
Iϵ(s) as follows:
Iϵ(s) = Z+∞
1
dx ϵ(x)
x1+s(31)
The function x→ϵ(x)
x1+sis integrable on [1,+∞) and thanks to the
integration by parts, we can write the following:
Iϵ(s) = Z+∞
1
dx ϵ(x)
x1+s(32)
=hΨ(x)
x1+si+∞
x=1 + (1 + s)Z+∞
1
dx Ψ(x)
x2+s(33)
= (1 + s)Z+∞
1
dx Ψ(x)
x2+s(34)
Since, sis a zeta zero, from the equation (6), we have:
Iϵ(s) = 1
s−1(35)
4

Therefore
Z+∞
1
dx Ψ(x)
x2+s=1
(s−1)(s+ 1) (36)
Thanks to equation (23), we can write:
Z+∞
1
dx ϕ(x)
x2+s=Z+∞
1
dx (x−1)
x2+s−2Z+∞
1
dx Ψ(x)
x2+s(37)
=1
s−1
s+ 1 −2
(s−1)(s+ 1) (38)
=1
s(1 −s)(39)
■
Lemma 1.3. Let us consider two variables σand τsuch that 0< σ < 1,
τ > 0and s=σ+iτis a zeta zero. Let us define the sequence of functions
ϕnand ψnover [0,+∞)and ϕnover [1,+∞)such that ϕ0(x) = ϕ0(x) =
ϕ(x)and for each n≥1,ϕn(0) = 0, and:
ϕn+1(x) = 1
xZx
0
dt ϕn(t)for x > 0 (40)
ϕn+1(x) = 1
xZx
1
dt ϕn(t)for x≥1 (41)
ψn(x) = x
2n−x2
3nfor x≥0 (42)
Therefore:
1. For each n≥0:
Z+∞
1
dx ϕn(x)
x2+s=1
s2n+1
(1 −s) 3n(43)
2. For each n≥0:
Z+∞
1
dx ϕn(x)
x3−s=1
(1 −s) 2n+1
s3n(44)
3. For each x≥1and n≥1:
ϕn(x) = ϕn(x) + 1
2nx
n−1
X
k=0
2klnk(x)
k!−1
3nx
n−1
X
k=0
3klnk(x)
k!(45)
4. For each n≥1and x≥1:
ϕn(x) = 1
(n−1)!
1
xZx
1
dt ϕ0(t)ln( x
t)n−1(46)
5

5. For each x≥0:
lim
n→+∞2nϕn(x) = x(47)
6. For each x≥0and n≥0:
0≤2nϕn(x)≤x(48)
7. For each x≥0and n≥1:
ψn(x) = ψn(x) + 1
2nx
n−1
X
k=0
2klnk(x)
k!−1
3nx
n−1
X
k=0
3klnk(x)
k!(49)
Where
ψn(x) = 1
(n−1)!
1
xZx
1
dt (t−t2)ln( x
t)n−1(50)
Proof. We have s=σ+iτazeta zero. The lemma 1.2 calculates the
integral A(s) as follows:
A(s) = Z+∞
1
dx ϕ0(x)
x2+s=1
s(1 −s)(51)
We use the integration by parts to write the following:
A(s) = Z+∞
1
dx ϕ0(x)
x2+s(52)
="1
x2+sZx
0
dx ϕ0(x)#+∞
1
+ (2 + s)Z+∞
1
dx ϕ1(x)
x2+s(53)
=−Z1
0
dx ϕ0(x) + (2 + s)Z+∞
1
dx ϕ1(x)
x2+s(54)
=−Z1
0
dx ϕ0(x)−(2 + s)Z1
0
dx ϕ1(x) + (2 + s)2Z+∞
1
dx ϕ2(x)
x2+s(55)
... (56)
=−
n
X
k=0
(2 + s)kZ1
0
dx ϕk(x) + (2 + s)n+1 Z+∞
1
dx ϕn+1(x)
x2+s(57)
We could do the above integration by parts because the functions
x→ϕk(x)
x2+sare piecewise continuous2and integrable over [1,+∞) as they
are dominated by the function x→1
x2+σthat is integrable over [1,+∞).
In fact, we can prove that the functions ϕkare non-negative and bounded
by 1 as we can prove by induction that for each kfor each x > 0 that:
0< ϕk+1(x) = 1
xZx
0
dt ϕk(t)≤1
xZx
0
1dt = 1 (58)
2All the functions x→ϕk(x)
x2+sare continuous except when k= 0 as the function ϕ0is
piecewise continuous.
6

This can be proven by induction using the fact that it is true for the initial
case as we have for each x≥0:
0≤ϕ0(x) = {x}(1 − {x})≤1 (59)
Now, we need to calculate the integrals Ikfor k≥0:
Ik=Z1
0
dx ϕk(x) (60)
For x∈(0,1), we have:
ϕ0(x) = x−x2(61)
And
ϕ1(x) = x
2−x2
3(62)
Therefore, we can write for each kfor x∈(0,1):
ϕk(x) = x
2k−x2
3k(63)
Therefore, for each k≥1:
Ik=1
2k+1 −1
3k+1 (64)
Therefore, we can conclude:
A(s) = 1
s(1 −s)+ (2 + s)n+1"Z+∞
1
dx ϕn+1(x)
x2+s− 1
s2n+1 +1
(1 −s) 3n+1 !#(65)
Since 2+s= 0, by combining the equations (51) and (65), we conculde
the first point of our lemma. ■
The second point of the lemma can be proved the same way like the
first point since 1 −sis also a zeta zero and 3 −s= 0(by applying the
first point on the zeta zero 1 −s). ■
The point 3) can be proved by induction. For n= 0. We have ϕ0(x) =
ϕ0(x). Let us assume that it is true till nand let us prove for n+ 1.
Therefore:
ϕn+1(x) = 1
xZ1
0
dt ϕn(t) + 1
xZx
1
dt ϕn(t) (66)
=1
x1
2n+1 −1
3n+1 +1
x2nZx
1
dt
t
n−1
X
k=0
2klnk(t)
k!(67)
−1
x3nZx
1
dt
t
n−1
X
k=0
3klnk(t)
k!+1
xZx
1
dt ϕn(t) (68)
=1
x1
2n+1 −1
3n+1 +1
x2n+1
n−1
X
k=0
2k+1 lnk+1(x)
(k+ 1)! (69)
−1
x3n+1
n−1
X
k=0
3k+1 lnk+1(x)
(k+ 1)! +ϕn+1(x) (70)
7

■
Let us prove the 4th point. We proceed by induction. For n= 1, we
retrieve the definition of ϕ1(x). Let us assume that it is true up to nand
let us prove it for n+ 1. Thanks to the integral order change, we can
write:
ϕn+1(x) = 1
xZx
1
dt ϕn(t) (71)
=1
(n−1)!
1
xZx
1
dt
tZt
1
ds ϕ0(s)ln( t
s)n−1(72)
=1
(n−1)!
1
xZx
1
ds ϕ0(s)Zx
s
dt
tln( t
s)n−1(73)
=1
n!
1
xZx
1
ds ϕ0(s)ln( x
s)n(74)
And this proves our point of the lemma. ■
Let us now prove the 5th point. Let us take x≥1. Thanks to d’Alembert’s
criterion, we have for each s∈[1, x], limn→+∞
2nln( x
s)n
n!= 0. We use
the formula of ϕn(x) in the previous point (4), and apply the dominated
convergence theorem to prove that:
lim
n→+∞2nϕn(x) = 1
xZx
1
lim
n→+∞
2nln( x
s)n
n!ϕ0(s)ds = 0 (75)
From point (3), we can conclude that:
lim
n→+∞2nϕn(x) = lim
n→+∞ 2nϕn(x) + 1
x"n−1
X
k=0
2klnk(x)
k!−2
3n
n−1
X
k=0
3klnk(x)
k!#!(76)
= lim
n→+∞
1
x"n−1
X
k=0
lnk(x2)
k!−2
3n
n−1
X
k=0
lnk(x3)
k!#(77)
=1
xexp(ln(x2)) = x(78)
For 0 ≤x < 1, we have from equation (63), ϕn(x) = x
2n−x2
3n, for each
n≥0. Therefore limn→+∞2nϕn(x) = x. Hence the proof of the point
(5). Let us now prove the point (6). We have for each x > 0:
0≤ϕ0(x)≤x(79)
Therefore by integrating the equation above we get:
0≤1
xZx
0
dt ϕ0(t)≤1
xZx
0
dt t (80)
Therefore
0≤ϕ1(x)≤x
2(81)
By induction, we easily conclude that for each n≥1:
0≤ϕn(x)≤x
2n(82)
8

■
Let us now prove the last point. We use Taylor’s Theorem with integral
form of remainder applied on the exponential function x→ex.
For x≥0 and n≥1:
ex=
n−1
X
k=0
xk
k!+1
(n−1)! Zx
0
dt etx−tn−1(83)
So, let us take x≥1. We can write:
eln(x2)=
n−1
X
k=0 ln(x2)k
k!+1
(n−1)! Zln(x2)
0
dt etln(x2)−tn−1(84)
We do a change of variable and write for ln(x2):
x2=
n−1
X
k=0
2klnk(x)
k!+2n
(n−1)! Zx
1
dt tln( x
t)n−1(85)
Therefore
x
2n=1
x2n
n−1
X
k=0
2klnk(x)
k!+1
x(n−1)! Zx
1
dt tln( x
t)n−1(86)
And the same for ln(x3):
x2
3n=1
x3n
n−1
X
k=0
3klnk(x)
k!+1
x(n−1)! Zx
1
dt t2ln( x
t)n−1(87)
Therefore
ψn(x) = x
2n−x2
3n=1
2nx
n−1
X
k=0
2klnk(x)
k!−1
3nx
n−1
X
k=0
3klnk(x)
k!+1
x(n−1)! Zx
1
dt (t−t2)ln( x
t)n−1
| {z }
ψn(x)
(88)
■
Lemma 1.4. Let us consider two variables σand τsuch that 0< σ < 1
2
and τ > 0such that s=σ+iτis a zeta zero. Let us define the sequence
of functions En,σ(x),Fn,σ (x)and Gn,σ (x)over [1,+∞)for each n≥0
as follows:
En,σ(x) = Zx
1
dt cos(τln (t))
t2+σϕn(t) (89)
Fn,σ(x) = Zx
1
dt sin(τln (t))
t2+σϕn(t) (90)
Gn,σ(x) = En,σ (x)Fn,1−σ(x)−En,1−σ(x)Fn,σ (x) (91)
1. There exists a0>1, there exists an integer n0such that for each
n≥n0:
Gn,σ(a0)>0 (92)
9

2. For large enough nthere exists xn> a0such as:
Gn,σ(xn) = 0 (93)
3. If the sequence (xn)is unbounded then:
σ=1
2(94)
4. If the sequence (xn)is bounded then:
σ=1
2(95)
Proof. Let us prove the first point. Let us prove it by contradiction. So
let us assume that the opposite is true. Therefore for each a0>1, for
each integer n0, there exists n≥n0such that:
Gn,σ(a0)≤0 (96)
So, let us take a0>1. Therefore for each n≥1, there exists kn≥nsuch
that:
Gkn,σ(a0+1
n)≤0 (97)
By construction we have:
lim
n→+∞kn= +∞(98)
We apply the dominated convergence theorem on the sequences of func-
tions:
f1
n(x) = cos(τln (t))
t2+σ2nϕn(t) (99)
f2
n(x) = sin(τln (t))
t2+σ2nϕn(t) (100)
f3
n(x) = cos(τln (t))
t3−σ2nϕn(t) (101)
f4
n(x) = sin(τln (t))
t3−σ2nϕn(t) (102)
From lemma 1.3, we have:
lim
n→+∞2nϕn(x) = x(103)
Therefore for each x≥1:
lim
n→+∞f1
kn(x) = f1(x) = cos(τln (x))
x1+σ(104)
lim
n→+∞f2
kn(x) = f2(x) = sin(τln (x))
x1+σ(105)
lim
n→+∞f3
kn(x) = f3(x) = cos(τln (x))
x2−σ(106)
lim
n→+∞f4
kn(x) = f4(x) = sin(τln (x))
x2−σ(107)
10

Thanks to the Dominated Convergence Theorem applied on the sequence
of functions in (99 −102) over the interval [1, a0], we apply the limit to
both sides of the equation(97) as follows:
lim
n→+∞Za0+1
n
1
dt f 1
n(x)Za0+1
n
1
dt f 4
n(x)≤lim
n→+∞Za0+1
n
1
dt f 2
n(x)Za0+1
n
1
dt f 3
n(x)(108)
Za0
1
dt cos(τln (t))
t1+σZa0
1
dt sin(τln (t))
t2−σ≤Za0
1
dt cos(τln (t))
t2−σZa0
1
dt sin(τln (t))
t1+σ(109)
From the lemma 1.1 result, we conclude:
g(a0) = J(a0,1 + σ, τ )I(a0,2−σ, τ )−J(a0,2−σ, τ)I(a0,1 + σ, τ )≤0 (110)
Where the function gis defined as follows:
g(x) = K(1 + σ, τ )K(2 −σ, τ ) 1 + σ(1 −σ)
τ2sin(τln (x))
xσ1−1
x1−2σ(111)
+(1 −2σ)
τ"cos(τln (x))
xσ1 + 1
x1−2σ−1 + 1
x#! (112)
Since a0is arbitrary, therefore for each x≥1:
g(x)≤0 (113)
Which is a contradiction since the function goscillates between negative
and positive values over [1,+∞). In fact, from lemma 2.4 (see Appendix),
there exists a constant τ0∼1.9208 such that for each τ > τ0, we have
g(e5π
2τ)>0 for each σ < 1
2. And for τ≤τ0, there is no zeta zero.
Therefore there is a0such that g(a0)>0 and hence the first point of
the lemma is proved. ■
Let us now prove the 2nd point of the lemma. From the lemma 1.3, thanks
to the results in (43) and (44), we can write:
lim
x→+∞Gn,σ(x) = Z+∞
1
dt cos(τln (t))
t2+σϕn(t)Z+∞
1
dt sin(τln (t))
t3−σϕn(t)(114)
−Z+∞
1
dt cos(τln (t))
t3−σϕn(t)Z+∞
1
dt sin(τln (t))
t2+σϕn(t)(115)
= σ
2n∥s∥2+(1 −σ)
3n∥1−s∥2! τ
2n∥1−s∥2−τ
3n∥s∥2!(116)
− (1 −σ)
2n∥1−s∥2+σ
3n∥s∥2! τ
2n∥s∥2−τ
3n∥1−s∥2!(117)
Therefore, we can write for large enough n:
lim
x→+∞Gn,σ(x)∼ − τ(1 −2σ)
∥s(1 −s)∥2
1
4n<0 (118)
11

Since the function Gn,σ is continuous over [1,+∞). From (92) and
(118) and thanks to the Mean value theorem, we can conclude that there
exists an xn> a0such that:
Gn,σ(xn) = 0 (119)
■
Let us prove the 3rd point. So, let us assume the sequence (xn) is
unbounded. There exists a subsequence3(xλ(n)) that tends to infinity
+∞. Without loss of generality, we assume that the limit of the sequence
(xn) is +∞. From the equation (93), we can deduce that:
Zxn
1
dt cos(τln (t))
t2+σϕn(t)Zxn
1
dt sin(τln (t))
t3−σϕn(t)=Zxn
1
dt cos(τln (t))
t3−σϕn(t)Zxn
1
dt sin(τln (t))
t2+σϕn(t)(120)
We multiply the two side of the equation above by 4n, we can write:
Zxn
1
dt cos(τln (t))
t2+σ2nϕn(t)Zxn
1
dt sin(τln (t))
t3−σ2nϕn(t)=Zxn
1
dt cos(τln (t))
t3−σ2nϕn(t)Zxn
1
dt sin(τln (t))
t2+σ2nϕn(t)(121)
We apply the Dominated Convergence Theorem on the sequences of func-
tions f1
n, f 2
n, f 3
n, f 4
ndefined in the equations (99 −102) since we have the
dominance condition for the functions fi
1≤i≤4:
fi
n(x)1≤i≤2≤
1
t2+σ2nϕn(t)≤1
t1+σ(122)
fi
n(x)3≤i≤4≤
1
t3−σ2nϕn(t)≤1
t2−σ(123)
Thanks to the Dominated Convergence Theorem applied over the in-
terval [1,+∞), we apply the limit to both sides of the equation(121) as
follows:
lim
n→+∞Zxn
1
dt f 1
n(x)Zxn
1
dt f 4
n(x)= lim
n→+∞Zxn
1
dt f 2
n(x)Zxn
1
dt f 3
n(x)(124)
Therefore
Z+∞
1
dt lim
n→+∞f1
n(x)Z+∞
1
dt lim
n→+∞f4
n(x)= lim
n→+∞Z+∞
1
dt lim
n→+∞f2
n(x)Z+∞
1
dt lim
n→+∞f3
n(x)(125)
Therefore
Z+∞
1
dt cos(τln (t))
t1+σZ+∞
1
dt sin(τln (t))
t2−σ=Z+∞
1
dt cos(τln (t))
t2−σZ+∞
1
dt sin(τln (t))
t1+σ(126)
From the lemma 1.1 result, we conclude:
σ
τK(1 + σ, τ )K(2 −σ, τ ) = 1−σ
τK(2 −σ, τ )K(1 + σ, τ ) (127)
Therefore:
σ= 1 −σ=1
2(128)
3Similar to Bolzano–Weierstrass theorem in the case of a bounded sequence.
12

■
Let us now prove the last point of the lemma. So, let us assume the se-
quence (xn) is bounded by an upper bound A > 0. Thanks to Bolzano–Weierstrass
theorem, there exists a subsequence (xλ(n)) that converges to a finite limit
a≥a0>1. And without loss of generality, we can assume that the limit
of the sequence (xn) is a.
Let us define the functions g1,g2and g3as follows:
g1(x) = J(x, 1 + σ, τ )I(x, 2−σ, τ )−J(x, 2−σ, τ)I(x, 1 + σ, τ ) (129)
g2(x) = J(x, 1 + σ, τ )I(x, 1−σ, τ ) + J(x, σ, τ)I(x, 2−σ, τ ) (130)
−J(x, 2−σ, τ )I(x, σ, τ )−J(x, 1−σ, τ)I(x, 1 + σ, τ ) (131)
g3(x) = J(x, σ, τ )I(x, 1−σ, τ )−J(x, 1−σ, τ)I(x, σ, τ ) (132)
From the lemma 1.3 point 3) and point 7), we can write for each x≥1:
ϕn(x) = ψn(x) + ϕn(x)−ψn(x)
| {z }
δn(x)
(133)
Where
ψn(x) = x
2n−x2
3n(134)
δn(x) = ϕn(x)−ψn(x) = 1
(n−1)!
1
xZx
1
dt ϕ0(t) + t2−tln( x
t)n−1(135)
Therefore for each x∈[1, A]:
|δn(x)| ≤ A2
(n−1)! ln(x)n−1(x−1)
x≤A2ln(A)n−1
(n−1)! (136)
We inject the equations (133 −134) into the equation (120) to write
the following. For large enough n:
Zxn
1
dt (ψn(x) + δn(x)f1(x)
xZxn
1
dt (ψn(x) + δn(x)f4(x)
x=Zxn
1
dt (ψn(x) + δn(x)f2(x)
xZxn
1
dt (ψn(x) + δn(x)f3(x)
x(137)
Where the functions fi1≤i≤4are defined in (104 −107). Therefore:
Zxn
1
dt ψn(x)f1(x)
xZxn
1
dt ψn(x)f4(x)
x−Zxn
1
dt ψn(x)f2(x)
xZxn
1
dt ψn(x)f3(x)
x(138)
=Zxn
1
dt δn(x)f2(x)
xZxn
1
dt ϕn(x)f3(x)
x+Zxn
1
dt δn(x)f3(x)
xZxn
1
dt ϕn(x)f2(x)
x(139)
−Zxn
1
dt δn(x)f1(x)
xZxn
1
dt ϕn(x)f4(x)
x−Zxn
1
dt δn(x)f4(x)
xZxn
1
dt ϕn(x)f1(x)
x(140)
Thanks to (136), from the equations (138-140) we can write:
Zxn
1
dt ψn(x)f1(x)
xZxn
1
dt ψn(x)f4(x)
x−Zxn
1
dt ψn(x)f2(x)
xZxn
1
dt ψn(x)f3(x)
x=O(ln(A)n−1
(n−1)! ) (141)
We inject the equation (134) into the equation (141). After simplification,
we can write the following:
1
4ng1(xn)−1
6ng2(xn) + 1
9ng3(xn) = O(ln(A)n−1
(n−1)! ) (142)
13

Therefore
3
22ng1(xn)−3
2ng2(xn) + g3(xn) = O(9 ln(A)n−1
(n−1)! ) (143)
Since the limit of the sequence ( 9 ln(A)n−1
(n−1)! ) is zero thanks to Stirling’s
formula. Since the functions g1,g2and g3are continuous and therefore
bounded over the interval [1, A]. There exists n0such that for each n≥n0,
we have:
g1(xn) = 0 (144)
g2(xn) = 0 (145)
Therefore
g1(a) = g2(a) = g3(a) = 0 (146)
Since a≥a0>1, hence σmust be equal to 1 −σ.
Therefore σ=1
2. This ends the proof of the Riemann Hypothesis. ■
1.4 Conclusion
We saw that if sis a zeta zero, then real part ℜ(s) can only be 1
2. Therefore
the Riemann’s Hypothesis is true: The non-trivial zeros of ζ(s)have real
part equal to 1
2. In the next article, we will apply the same method to
prove the Generalized Riemann Hypothesis (GRH).
Acknowledgments
I would like to thank Farhat Latrach, Giampiero Esposito, Jacques G´elinas,
Michael Milgram, L´eo Ag´elas, Ronald F. Fox, Kim Y.G, Masumi Naka-
jima, Maksym Radziwill and Shekhar Suman for thoughtful comments
and discussions on my paper versions on the RH. All errors are mine.
2 Appendix
Let us denote the following functions:
cos+(x) = max cos(x),0(147)
cos−(x) = max −cos(x),0(148)
Lemma 2.1. Let us consider two variables σand τsuch that σ > 0and
τ > 0. Let us define two integrals I(+, σ, τ )and I(−, σ, τ )as follows:
I(+, σ, τ ) = Z+∞
1
cos+(τln(x))
x1+σdx (149)
I(−, σ, τ ) = Z+∞
1
cos−(τln(x))
x1+σdx (150)
14

Therefore
I(+, σ, τ ) = K(1 + σ, τ )σ
τ+K(1 + σ, τ )e−πσ
2τ
1−e−πσ
τ
(151)
I(−, σ, τ ) = K(1 + σ, τ )e−πσ
2τ
1−e−πσ
τ
(152)
Proof. Let us calculate I(+, σ, τ ) as follows:
I(+, σ, τ ) = Z+∞
1
dx cos+(τln (x))
x1+σ(153)
=Ze
π
2τ
1
dx cos(τln (x))
x1+σ+
+∞
X
k=0 Ze
(4k+5)π
2τ
e
(4k+3)π
2τ
dx cos(τln x)
x1+σ(154)
=Ze
π
2τ
1
dx cos(τln (x))
x1+σ+
+∞
X
k=0
I(k, σ, τ ) (155)
Where
I(k, σ, τ ) = Ze
(4k+5)π
2τ
e
(4k+3)π
2τ
dx cos(τln x)
x1+σ(156)
From lemma 1.1, we can write the following:
I(k, σ, τ ) = K(1 + σ, τ) e−(4k+5)πσ
2τ+e−(4k+3)πσ
2τ!(157)
Therefore
I(+, σ, τ ) = K(1 + σ, τ ) σ
τ+e−πσ
2τ!+K(1 + σ, τ )
+∞
X
k=0 e−(4k+5)πσ
2τ+e−(4k+3)πσ
2τ!(158)
=K(1 + σ, τ ) σ
τ+e−πσ
2τ!+K(1 + σ, τ )
+∞
X
k=0 e−5πσ
2τe−2kπσ
τ+e−3πσ
2τe−2kπσ
τ!(159)
=K(1 + σ, τ ) σ
τ+e−πσ
2τ!+K(1 + σ, τ )e−5πσ
2τ+e−3πσ
2τ
1−e−2πσ
τ
(160)
=K(1 + σ, τ ) σ
τ+e−πσ
2τ!+K(1 + σ, τ )e−3πσ
2τ1 + e−πσ
τ
1−e−2πσ
τ
(161)
=K(1 + σ, τ ) σ
τ+e−πσ
2τ!+K(1 + σ, τ )e−3πσ
2τ1
1−e−πσ
τ
(162)
=K(1 + σ, τ )σ
τ+K(1 + σ, τ )e−πσ
2τ 1 + e−πσ
τ
1−e−πσ
τ!(163)
=K(1 + σ, τ )σ
τ+K(1 + σ, τ )e−πσ
2τ
1−e−πσ
τ
(164)
And therefore
15

I(−, σ, τ ) = I(+∞, σ, τ )−I(+, σ, τ) (165)
=K(1 + σ, τ )e−πσ
2τ
1−e−πσ
τ
(166)
■
Lemma 2.2. Let us consider fa continuous function over [1,+∞). let’s
ϕbe a non-null positive function such that fϕ and ϕare integrable func-
tions over [1,+∞)with:
0<Z+∞
1
dx ϕ(x)<+∞(167)
And
Z+∞
1
dx ϕ(x)f(x)<+∞(168)
Therefore, there exists a real c∈(1,+∞)such that:
Z+∞
1
dx ϕ(x)f(x) = f(c)Z+∞
1
dx ϕ(x) (169)
Proof. Let us define the real λas following:
λ=R+∞
1dx ϕ(x)f(x)
R+∞
1dx ϕ(x)(170)
We have by construction that:
Z+∞
1
dx ϕ(x)f(x)−λ= 0 (171)
Therefore, if for each x > 1, we have f(x)> λ, then,we will have:
Z+∞
1
dx ϕ(x)f(x)−λ>0 (172)
Which is a contradiction. We will reach a similar contradiction if we
assume f(x)< λ for each x > 1. Therefore, there exists c∈(1,+∞) such
that f(c) = λ.■
Lemma 2.3. Let us consider two variables σand τsuch that 0< σ < 1
and τ > 0and s=σ+iτis a zeta zero. Therefore:
τ > √3π
2s1 + 1
1 + 1
2pπ
3∼1.9786 (173)
16

Proof. From the lemma 1.3, we have:
Z+∞
1
dx ϕ(x) cos(τln (x))
x2+σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(174)
Therefore, we can write the following:
Z+∞
1
dx ϕ(x) cos(τln (x))
x2+σ=Z+∞
1
dx ϕ(x) cos+(τln (x))
x2+σ−Z+∞
1
dx ϕ(x) cos−(τln (x))
x2+σ(175)
The function x→ϕ(x) is continuous over [0,+∞). Therefore, from
lemma 2.2, there exists c1>1 and c2>1 such that:
Z+∞
1
dx ϕ(x) cos(τln (x))
x2+σ=ϕ(c1)
c2σ
1Z+∞
1
dx cos+(τln (x))
x2−σ−ϕ(c2)
c2σ
2Z+∞
1
dx cos−(τln (x))
x2−σ(176)
Let us denote α1=ϕ(c1)
c2σ
1and α2=ϕ(c2)
c2σ
2. We have 0 < α1<1
4and
0< α2<1
4.
Case 1: α1<=α2In this case we can write α1=α2−ϵwith
0< ϵ < 1
4. Therefore we can write from the equation (174) that:
α1Z+∞
1
dx cos+(τln (x))
x2−σ−α2Z+∞
1
dx cos−(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(177)
Therefore
α1Z+∞
1
dx cos+(τln (x))
x2−σ−α1Z+∞
1
dx cos−(τln (x))
x2−σ−ϵZ+∞
1
dx cos−(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(178)
Therefore
α1Z+∞
1
dx cos(τln (x))
x2−σ−ϵZ+∞
1
dx cos−(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(179)
Therefore
α11−σ
(1 −σ)2+τ2−ϵZ+∞
1
dx cos−(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(180)
Therefore
−ϵZ+∞
1
dx cos−(τln (x))
x2−σ=σ
σ2+τ2+ (1 −α1)1−σ
(1 −σ)2+τ2(181)
Which is a contradiction.
17

Case 2: α1> α2In this case we can write α1=α2+ϵwith 0 < ϵ < 1
4.
Therefore we can write from the equation (174) that:
α1Z+∞
1
dx cos+(τln (x))
x2−σ−α2Z+∞
1
dx cos−(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(182)
Therefore
α2Z+∞
1
dx cos+(τln (x))
x2−σ−α2Z+∞
1
dx cos−(τln (x))
x2−σ+ϵZ+∞
1
dx cos+(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(183)
Therefore
α2Z+∞
1
dx cos(τln (x))
x2−σ+ϵZ+∞
1
dx cos+(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(184)
Therefore
α21−σ
(1 −σ)2+τ2+ϵZ+∞
1
dx cos+(τln (x))
x2−σ=σ
σ2+τ2+1−σ
(1 −σ)2+τ2(185)
Therefore
+ϵZ+∞
1
dx cos+(τln (x))
x2−σ=σ
σ2+τ2+ (1 −α2)1−σ
(1 −σ)2+τ2(186)
From lemma 2.1, we have:
Z+∞
1
dx cos+(τln (x))
x2−σ=1−σ
(1 −σ)2+τ2+τ
(1 −σ)2+τ2
e−π(1−σ)
2τ
1−e−π(1−σ)
τ
(187)
Therefore
ϵ τ
(1 −σ)2+τ2
e−π(1−σ)
2τ
1−e−π(1−σ)
τ
=σ
σ2+τ2+ (1 −α1)1−σ
(1 −σ)2+τ2(188)
Therefore
ϵ τ e−π(1−σ)
2τ
1−e−π(1−σ)
τ
=σ(1 −σ)2+τ2
σ2+τ2+ (1 −α1)1−σ(189)
Therefore
ϵ τ e−π(1−σ)
2τ
1−e−π(1−σ)
τ
=σ(1 −2σ)
σ2+τ2+σ+1−σ−α11−σ(190)
We have 0 < σ < 1
2. Therefore
ϵ τ e−π(1−σ)
2τ
1−e−π(1−σ)
τ
>1−α11−σ(191)
18

We have 0 < α1<1
4and 1 >1−σ > 1
2. Therefore: 1−α11−σ>3
4.
Therefore
ϵ τ e−π(1−σ)
2τ
1−e−π(1−σ)
τ
>3
4(192)
But we also have 0 < ϵ < 1
4. Therefore
3
4<τ
4
e−π(1−σ)
2τ
1−e−π(1−σ)
τ
(193)
Since ex≥1 + xfor each x≥0, therefore:
τ
3> e
π(1−σ)
2τ−e−π(1−σ)
2τ(194)
>1 + π(1 −σ)
2τ−1
1 + π(1−σ)
2τ
=π(1 −σ)
2τ+
π(1−σ)
2τ
1 + π(1−σ)
2τ
(195)
Since the function x→x
1+xis increasing over [0,+∞) and π(1−σ)
2τ≥π
4τ,
we can write:
4τ2
3π>1 + 4τ
4τ+π(196)
Since τ > 0, therefore,
τ > √3π
2(197)
Since the function x→x
x+πis increasing over [0,+∞). Therefore
4τ2
3π>1 + 2√3π
2√3π+π(198)
Therefore
τ > √3π
2s1 + 1
1 + 1
2pπ
3
(199)
Remark. We can reiterate the same procedure using the equation (196)
to further improve the minimum bound. The limit bound τmin is actually
the root of the equation 16
3πτ3+4
3τ2−8τ−π= 0.τmin ∼2.01271781.■
Lemma 2.4. Let us consider two variables σand τsuch that 0< σ <
1and τ > 0. Let us consider the function gdefined in the equation
(111 −112). Therefore there exists a constant τ0∼1.9208 such that:
1. For τ > τ0, we have g(e5π
2τ)>0for each σ < 1
2.
2. For τ≤τ0,s=σ+iτcannot be a zeta zero.
19

Figure 1: Functions f5
2(x, σ) for different values of σfrom 0 to 1
2and function
h5
2(x) over [0,5] .
Proof. Let us prove the first point of the lemma. As we are interested
to the sign of the function g, we will ignore the positive term K(1 +
σ, τ )K(2 −σ, τ ) from the gexpression as follows:
g(x) = 1 + σ(1 −σ)
τ2sin(τln (x))
xσ1−1
x1−2σ(200)
+(1 −2σ)
τ"cos(τln (x))
xσ1 + 1
x1−2σ−1 + 1
x#(201)
We can find asuch that g(a)>0 of the form a=eλπ
τwhere λis of the
form λ(k) = 2k+1
2and k∈N. For such a, we have:
g(a) = 1 + σ(1 −σ)
τ2e−λπσ
τ1−e−λπ(1−2σ)
τ−(1 −2σ)
τ1 + e−λπ
τ(202)
Let us denote x=λπ
τand fλ(x) = g(ex). Therefore
fλ(x, σ) = 1 + σ(1 −σ)
λπ2x2e−xσ −e−x(1−σ)−(1 −2σ)
λπ x1 + e−x(203)
We can use the asymptotic expansion around 0 (x→0) of fλto get
an idea of what is happening around 0. For ϵ > 0 small enough, we can
write:
fλ(ϵ, σ) = 1−2σ1−2
λπ ϵ−1
2ϵ2+O(ϵ3) (204)
To get fλ(ϵ, σ)>0, we need λ≥2
πand therefore k≥1 since 1−2σ > 0.
For the rest of this section we will take λ=λ(1) = 5
2. We will plot the
function f5
2over [0,5] for different values of σfrom 0 to 1
2in figure 1.
20

From figure 1, it appears that there exists a constant a0such that for
each x∈(0, a0), we have f5
2(x)>0 for each σ < 1
2. We are going to
prove that this remark holds true. Also, the more σis close to 1
2, the
more fλ(a0, σ) is close to zero. The asymptotic expansion of fλwhen σ
tends to 1
2is as follows:
fλ(x, σ) = 1−2σxe−x
21 + x2
(2λπ)2−ex
2+e−x
2
λπ 
| {z }
hλ(x)
+O((σ−1
2)3) (205)
The function hλis continous over [0,+∞). The asymptotic expansion of
hλwhen xtends to 0+is as follows:
hλ(x) = 1−2
λπ x+O(x2) (206)
Therefore hλ(x)∼x→0+1−2
λπ x > 0. The limit of hλ(x) when xgoes
to infinity is −∞. Hence, thanks to the mean value theorem, the function
hλhas at least one non-null zero. Let’s x0be the first non-null zero of the
function hλ. Hence, for each x∈[0, x0], hλ(x)≥0. Numerically, Newton-
Raphson method locates the value of x0in the interval (4.233,4.234). Now
let us take x∈[0, x0]. From the equation (203), we can write:
fλ(x, σ) = e−x
21 + σ(1 −σ)
λπ2x2ex(1
2−σ)−e−x(1
2−σ)−(1 −2σ)
λπ x1 + e−x(207)
= 2e−x
21 + σ(1 −σ)
λπ2x2sinh(x(1
2−σ)) −(1 −2σ)
λπ x1 + e−x(208)
Since hλ(x)≥0, we have:
x2
(2λπ)2≥ex
2+e−x
2
λπ −1 (209)
Also, we have for t≥0, sinh(t)≥t, therefore we can write:
fλ(x, σ)≥2e−x
21 + σ(1 −σ)
λπ2x2x(1
2−σ)−(1 −2σ)
λπ x1 + e−x(210)
≥(1 −2σ)x"e−x
21 + 4σ(1 −σ)x2
(2λπ)2−1 + e−x
λπ #(211)
≥(1 −2σ)x"e−x
2+ 4σ(1 −σ)h1 + e−x
λπ −e−x
2i−1 + e−x
λπ #(212)
≥(1 −2σ)1−4σ(1 −σ)xhe−x
2−1 + e−x
λπ i(213)
≥2(1 −2σ)1−4σ(1 −σ)xe−x
2
λπ hλπ
2−ex
2+e−x
2
2i(214)
≥2(1 −2σ)1−4σ(1 −σ)xe−x
2
λπ hλπ
2−cosh( x
2)i(215)
21

Let us denote x1= 2 arcosh λπ
2= 2 ln λπ
2+q(λπ
2)2−1. Numerically,
x1∼4.0888 and x1< x0. Therefore for each 0 < x < x1,fλ(x, σ)>0.
Therefore, there exists a constant τ0=5π
2x1∼1.9208 such that for
each τ > τ0, we have g(e5π
2τ)>0 for each σ < 1
2.■
Let us prove the second point of the lemma. Thanks to lemma 2.3,
we have τ > τ1=√3π
2q1 + 1
1+ 1
2√π
3∼1.9786. Therefore if τ≤τ0, then
τ < τ1, therefore s=σ+iτcannot be a zeta zero. ■
References
[1] Bernhard Riemann. On the Number of Prime Numbers less than a
Given Quantity
https://www.claymath.org/sites/default/files/ezeta.pdf
[2] Aleksandar Ivic. The Riemann Zeta-Function: Theory and Applica-
tions
[3] Peter Borwein, Stephen Choi, Brendan Rooney, and Andrea Weirath-
mueller The Riemann Hypothesis: A Resource for the Afficionado
and Virtuoso Alike
http://wayback.cecm.sfu.ca/~pborwein/TEMP_PROTECTED/book.
pdf
[4] Jørgen Veisdal. The Riemann Hypothesis, explained
https://medium.com/cantors-paradise/
the-riemann-hypothesis-explained- fa01c1f75d3f
[5] Thai Pham. Dirichlet’s Theorem on Arithmetic Progressions
https://web.stanford.edu/ thaipham/papers/MIT 18.104 Review Paper.pdf
[6] G. H. Hardy. The general theory of dirichlet series.
https://archive.org/details/generaltheoryofd029816mbp/page/n9
[7] Garrett, Paul. Primes in arithmetic progressions, 2011.
http : //www.math.umn.edu/ garrett/m/mfms/notes c/dirichlet.pdf
[8] Eissa D. Habil. Double Sequences and Double Series.
https://journals.iugaza.edu.ps/index.php/IUGNS/article/download/1594/1525
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