RH_Presentation.2022.1.pdf

A Simple Proof of the Riemann Hypothesis
Charaf Ech-Chatbi
charaf@melix.net
May 26, 2022
Charaf Ech-Chatbi Riemann Hypothesis May 26, 2022 1 / 29

Overview
1 Riemann Hypothesis
The importance of RH
Riemann Zeta Function
2 The proof of RH
Introduction
Lemma 1.1
Lemma 1.2
Lemma 1.4
The Proof

Riemann Hypothesis The importance of RH
The importance of RH
The prime number theorem determines the average distribution of the
primes:
π(x) ∼
x→+∞
x
ln(x)
(1)
where π(x) is the prime counting function:
π(x) = the number of primes ≤ x (2)
Formulated by both Gauss and Legendre independently, the prime
number theorem states that for a great x → +∞, the prime counting
function π(x) will approximate the function x
ln(x) .

The importance of RH (Cont)
Formulated in Riemann’s 1859 paper, Riemann hypothesis asserts that
all the ’non-trivial’ zeros of the Zeta function are complex numbers
with real part 1
2. In the same paper, Riemann gave us a formula that
explicitly tells us about the deviation from the average x
ln(x) :
J(x) = Li(x)
| {z }
Term 1
−
X
ρ
Li(xρ
)
| {z }
Term 2
− ln(2) +
Z +∞
x
dt
t(t2 − 1) ln(t)
| {z }
Term 3
(3)
Where Li(x) is the logarithmic integral function:
Li(x) =
Z x
2
dt
ln(t)
(4)
∼
x→+∞
x
ln(x)
(5)

Term 1 is the logarithmic integral Li(x). It is the estimation of π(x)
in the prime number theorem. It is the dominant term in the J(x)
formula. It is actually an overestimation of π(x).
Term 2 is the sum of Li(xρ) over the non-trivial zeros ρ of the Zeta
function. It adjusts the overestimation of Term 1.
Term 3 is the integral term minus ln(2).
J(x) is the Riemann prime counting function:
J(x) =
+∞
X
k=1
1
k
π(x
1
k ) (6)

And its Mobius inversion
π(x) =
+∞
X
k=1
µ(k)
k
J(x
1
k ) (7)
Notice that J(x) and π(x) are finite sums. At some term k, we will
have x
1
k < 2. And the Mobius function µ(n):
µ(n) =





1, if n is square-free with an even number of prime factors
0, if n is not square-free
−1, if n is square-free with an odd number of prime factors
(8)

In summary: RH tells us about the deviation from the average
distribution of the primes.
Professor Peter Sarnak has suggested in a lecture, available online,
that there are probably ”500 or more” theorems that depend on RH
or GRH.

Riemann Hypothesis Riemann Zeta Function
Riemann Zeta Function
For a complex number s where <(s) > 1, the Zeta function is defined
as the sum of the series:
ζ(s) =
+∞
X
n=1
1
ns
(9)
In his 1859 paper, Riemann extended the zeta function ζ(s), by
analytical continuation, to an absolutely convergent function in the
half plane <(s) > 0, minus a simple pole at s = 1:
ζ(s) =
s
s − 1
− s
Z +∞
1
{x}
xs+1
dx (10)
Where {x} = x − [x] is the fractional part and [x] is the integer part
of x.
Riemann obtained the analytic continuation of the zeta function to
the whole complex plane.

Riemann Zeta Function (Cont)
Riemann has shown that Zeta has a functional equation
ζ(s) = 2s
πs−1
sin
πs
2

Γ(1 − s)ζ(1 − s) (11)
Where Γ(s) is the Gamma function.
Riemann has shown that the non-trivial zeros of ζ are located
symmetrically with respect to the line (s) = 1/2, inside the critical
strip 0 (s) 1.
Riemann has conjectured that all the non trivial-zeros are located on
the critical line (s) = 1/2.
In 1921, Hardy Littlewood showed that there are infinitely many
zeros on the critical line.
In 1896, Hadamard and De la Vallée Poussin proved that ζ(s) has no
zeros of the form s = 1 + it for t ∈ R.

Riemann Zeta Properties
Some of the known results of ζ(s) are:
ζ(s) has no zero for (s) 1.
ζ(s) has no zero of the form s = 1 + iτ. i.e. ζ(1 + iτ) 6= 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 (s) 1.
If ζ(s) = 0, then 1 − s, s̄ and 1 − s̄ are also zeros of ζ: i.e.
ζ(s) = ζ(1 − s) = ζ(s̄) = ζ(1 − s̄) = 0.
To prove the RH, it is sufficient to prove that ζ has no zero on
0 (s) 1
2 of the left side of the critical strip.

The proof of RH Introduction
Proof of the Riemann Hypothesis
“Without doubt it would be desirable to have a rigorous proof of
this proposition; however I have left this research aside for the
time being after some quick unsuccessful attempts, because it
appears to be unnecessary for the immediate goal of my
investigation.”
– Bernhard Riemann 1859

The proof of RH Introduction
Proof of the Riemann Hypothesis (Cont)
Let’s take a complex number s such that s = σ + iτ.
Let’s suppose that 0 σ 1, τ 0 and ζ(s) = 0.
We have the Riemann’s integral:
ζ(s) =
s
s − 1
− s
Z +∞
1
{x}
xs+1
dx (12)
We have s 6= 1, s 6= 0 and ζ(s) = 0, therefore:
1
s − 1
=
Z +∞
1
{x}
xs+1
dx (13)
Let’s denote some functions:
(x) = {x}, φ(x) = {x} 1 − {x}

, Ψ(x) =
Z x
1
dx (x) (14)
To continue, we will need some few lemmas.

The proof of RH Lemma 1.1
Proof - Lemma 1.1
The function (x) is piecewise continuous and its primitive function Ψ(x)
is defined as follows:
Ψ(x) =
1
2

x − 1 − φ(x)

(15)
Let’s consider two variables σ and τ such that 0 σ 1 and τ 0 such
that s = σ + iτ is a zeta zero. Therefore:
Z +∞
1
dx
Ψ(x)
x2+s
=
1
(s − 1)(1 + s)
(16)
Z +∞
1
dx
φ(x)
x2+s
=
1
s(1 − s)
(17)

Proof - Lemma 1.2
Let’s consider two variables σ and τ such that 0 σ 1, τ 0 and
s = σ + iτ is a zeta zero. Let’s define the sequence of functions ψn such
that ψ0(x) = φ(x) = {x} 1 − {x}

and for each n ≥ 1:
ψn+1(x) =
1
x
Z x
0
ψn(t) (18)
Therefore:
For each n ≥ 1:
Z +∞
1
dx
ψn(x)
x2+s
=
1
s 2n
+
1
(1 − s) 3n
(19)
For each n:
1
(n − 1)!
Z +∞
1
dx
R x
1 dt ψ0(t) ln(x
t )
n−1
x3+s
=
1
s(1 − s) (2 + s)n
(20)

Proof - Lemma 1.4
Let’s consider two variables σ and τ such that 0 σ ≤ 1
2, τ 0 and
s = σ + iτ is a zeta zero. Let’s define the sequence of functions ψ̃n such
that ψ̃0(x) = ψ0(x)
x1−2σ =
{x} 1−{x}

x1−2σ and for each n ≥ 1:
ψ̃n+1(x) =
1
x
Z x
0
ψ̃n(t) (21)
Therefore:
For each n:
Z +∞
1
dx
ψ̃n(x)
x2+s
=
1
(1 − s) (1 + 2σ)n
+
1
s (2 + 2σ)n
(22)
Where s is the complex conjugate of s.
For each n:
1
(n − 1)!
Z +∞
1
dx
R x
1 dt ψ̃0(t) ln(x
t )
n−1
x3+s
=
1
s(1 − s) (2 + s)n
(23)

Proof - Lemma 1.4 cont

For each n:
Z +∞
1
dx
cos(τ ln (x))
R x
1 dt ψ̃0(t) ln(x
t )
n−1
x3+σ
(24)
=
1
c1−2σ
n
Z +∞
1
dx
2 cos2(τ
2 ln (x))
R x
1 dt ψ0(t) ln(x
t )
n−1
x3+σ
(25)
−
1
c1−2σ
Z +∞
1
dx
R x
1 dt ψ0(t) ln(x
t )
n−1
x3+σ
(26)
Where cn 1 and c 1 are real numbers.

Proof - Lemma 1.4 cont

Let’s define the sequence zn as follows:
zn =
cos(α − βn) − cos(α+βn)
c1−2σ
c1−2σ
n
(27)
Let’s assume α ∈ (0, π
2 ) and β ∈ (0, π
2 ). Let’s assume further that the
sequence (cn) converges to c but its terms are different from c such
that for each n ≥ 1, there exist an integer k ≥ n such that ck 6= c.
Therefore, we can always find an integer n such that zn 0.
Even if the sequence (cn) converges, the sequence (zn) does not
converge.

Proof - Summary
From the lemmas 1.2 and 1.4, we have for n ≥ 1:
1
(n − 1)!
Z +∞
1
dx
R x
1 dt ψ0(t) ln(x
t )
n−1
x3+s
=
1
s(1 − s) (2 + s)n
(28)
And
1
(n − 1)!
Z +∞
1
dx
R x
1 dt ψ̃0(t) ln(x
t )
n−1
x3+s
=
1
s(1 − s) (2 + s)n
(29)
Where s is the complex conjugate of s.

The proof of RH The Proof
Proof - Final line
Let’s write the following:
1
s(1 − s)
=
e−iα
ks(1 − s)k
(30)
1
2 + s
=
e−iβ
k2 + sk
(31)
Where α ∈ (0, π) and β ∈ (0, π). Without loss of generality, we can
assume α and β are in (0, π
2 ) since 0 σ 1
2 and τ 0.
Therefore, we can write:
1
(2 + s)n
=
e−iβn
k2 + skn
(32)

Proof - Final line (Cont)
Therefore, when we combine the equations (24-26) and( 28-29) and get
the following:
Xn =
1
(n − 1)!
Z +∞
1
dx
2 cos2(τ
2 ln (x))
R x
1 dt ψ0(t) ln(x
t )
n−1
x3+σ
(33)
=
1
ks(1 − s)kk2 + skn
c1−2σ
1
c1−2σ
n
− 1
c1−2σ
(34)
And
Yn =
1
(n − 1)!
Z +∞
1
dx
R x
1 dt ψ0(t) ln(x
t )
n−1
x3+σ
(35)
=
1
c1−2σ
n
1
c1−2σ
n
− 1
c1−2σ
(36)

We can also calculate Xn and Yn in another way as follows:
Xn =
U0
(2 + σ)n
+
cos(α + βn)
(37)
And
Yn =
U0
(2 + σ)n
(38)
Where
U0 =
Z +∞
1
dt
ψ0(t)
t2+σ
(39)
From the equations (34) and (37), we deduce that the limit of the
sequence ( 1
cn
)n≥1 is 1
c 1.

Case 1: ∃n0 ∈ N, ∀n ≥ n0: cn = c
We have from the equations (33-38) that for n ≥ n0:
Xn
Yn
= 1 +
(2 + σ)n cos(α + βn)
U0ks(1 − s)kk2 + skn
=
c1−2σ
c1−2σ
n
(40)
Therefore since cn = c, we can write:
1 +
(2 + σ)n cos(α + βn)
U0ks(1 − s)kk2 + skn
= 1 (41)
Since σ 0, therefore for each n ≥ n0:
cos(α + βn) = cos(α) cos(βn) − sin(α) sin(βn) = 0 (42)
Since β ∈ (0, π
2 ), therefore
sin(α) = cos(α) = 0 (43)
Which is a contradiction.

Case 2: ∀n0 ∈ N, ∃n ≥ n0: cn 6= c
We have from the equation (40):
Xn
Yn
=
c1−2σ
c1−2σ
n
(44)
Since 0 σ 1
2 and τ 0, we have α ∈ (0, π
2 ), β ∈ (0, π
2 ), sin(α) 6= 0
and cos(α) 6= 0. And since the sequence (cn) terms and c are different,
from the lemma 1.4, we can find n0 such that
cos(α − βn0) − cos(α+βn0)
c1−2σ
n0
0 and cos(α − βn0) − cos(α+βn0)
c1−2σ 0.
Therefore, in such case, we will have
Xn0
Yn0
0. Which means either
Xn0 0 or Yn0 0. Which is a contradiction.

Case 2: ∀n0 ∈ N, ∃n ≥ n0: cn 6= c (Cont)
We can also write the sequence (Xn
Yn
) as follows:
Xn
Yn
=
A cos(βn) + B sin(βn)
An cos(βn) + Bn sin(βn)
(45)
Where the sequences (An) and (Bn) converge respectively to the limits A
and B.
An = cos(α)

1 −
1
c1−2σ
n

, Bn = sin(α)

1 +
1
c1−2σ
n

(46)
A = cos(α)

1 −
1
c1−2σ

, B = sin(α)

1 +
1
c1−2σ

(47)

Case 2: ∀n0 ∈ N, ∃n ≥ n0: cn 6= c (Cont)
From the equation (40), the sequence (Xn
Yn
) converges to 1.
And therefore the sequence ( A cos(βn)+B sin(βn)
An cos(βn)+Bn sin(βn) ) converges to 1.
Since sin(α) 6= 0 and cos(α) 6= 0, the sequence ( A cos(βn)+B sin(βn)
An cos(βn)+Bn sin(βn))
does not converge.
Hence the contradiction.

Proof of RH
In constrast, in case α = 0 or α = π
2 , we will have Xn
Yn
=
1− 1
c1−2σ
1− 1
c1−2σ
n
always non-negative / converges.
And since sin(α) = τ(1−2σ)
σ(1−σ)+τ2
2
+(τ(1−2σ))2
1, α cannot be π
2 .
Therefore α should be zero and:
σ =
1
2
(48)
In case of σ 1
2. We work with 1 − s that is also a zeta zero and
1 − σ 1
2.
And we will reach the same conclusion. This ends the proof of the
Riemann Hypothesis.

Conclusion
The same method[2] can be applied to prove the Generalized
Riemann Hypothesis (GRH).
RH is true: The non-trivial zeros of ζ(s) have real part equal to 1
2.
GRH is true: The non-trivial zeros of a L − Dirichlet function have
real part equal to 1
2.

The proof of RH References
References
Charaf Ech-Chatbi (2022)
A Simple Proof of the Riemann’s Hypothesis
https://papers.ssrn.com/sol3/papers.cfm?abstract id=3390234
Charaf Ech-Chatbi (2022)
A Simple Proof of the Generalized Riemann Hypothesis
https://papers.ssrn.com/sol3/papers.cfm?abstract id=3398007

The proof of RH QED
The End

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