A Proof Of The Riemann Hypothesis Based On A New Expression Of The Completed Zeta Function

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A Proof Of The Riemann Hypothesis Based On A New Expression Of The Completed Zeta Function

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Weicun Zhang^*

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Submitted:

14 April 2023

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17 April 2023

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Abstract

Based on the Hadamard product $\xi(s)= \xi(0)\prod_{\rho}(1-\frac{s}{\rho})$, a new absolute convergent expression of $\xi(s)$ is obtained by paring $\rho_i$ and $\bar{\rho}_i$, and putting all the $\rho_i$ related multiple zeros together in one factor, i.e., $$\xi(s)=\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}}$$ where $\xi(0)=\frac{1}{2}$, $\rho_i=\alpha_i+j\beta_i$ and $\bar{\rho}_i=\alpha_i-j\beta_i$ are the complex conjugate zeros of $\xi(s)$, $0<\alpha_i<1$ and $\beta_i\neq 0$ are real numbers, $d_i\geq 1$ are the real multiplicities of $\rho_i$, $\beta_i$ are in order of increasing $|\beta_i|$, i.e., $|\beta_1|\leq|\beta_2|\leq|\beta_3|\leq \cdots$. Then, by the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}} =\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(1-s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}}$$ i.e., $$\prod_{i=1}^{\infty}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)}^{d_{i}}=\prod_{i=1}^{\infty}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}^{d_{i}}$$ which, by Lemma 3, is equivalent to $$\begin{cases}&\alpha_i=\frac{1}{2}, i =1,2,3, \cdots, \infty\\ & |\beta_1|<|\beta_2|<|\beta_3|< \cdots\\ \end{cases}$$ Thus, we conclude that the Riemann Hypothesis is true.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

It has been almost 164 years since the Riemann Hypothesis (RH) was proposed in 1859 [1]. Many efforts and achievements have been made towards proving this celebrated hypothesis, but it is still an open problem [2,3]. The Riemann zeta function is the function of the complex variable s, defined in the half-plane

ℜ (s) > 1

by the absolutely convergent series [2]

ζ (s) = \sum_{n = 1}^{\infty} \frac{1}{n^{s}}, ℜ (s) > 1

(1)

The connection between the Riemann zeta function and prime numbers can be established through the well-known Euler product, i.e.

ζ (s) = \sum_{n = 1}^{\infty} \frac{1}{n^{s}} = \prod_{p} {(1 - p^{- s})}^{- 1}, ℜ (s) > 1

(2)

where p runs over the prime numbers.

Riemann showed how to extend zeta function to the whole complex plane

C

by analytic continuation

ζ (s) = \frac{π^{s / 2}}{Γ (s / 2)} {\frac{1}{s (s - 1)} + \int_{1}^{\infty} (x^{\frac{s}{2} - 1} + x^{- \frac{s}{2} - \frac{1}{2}}) \cdot (\frac{θ (x) - 1}{2}) d x}

(3)

where

θ (x) = \sum_{- \infty}^{\infty} e^{- n^{2} π x}

being the Jaccobi theta function,

Γ

being the Gamma function in the following Weierstrass expression (Meanwhile, there are also Gauss expression, Euler expression, and integral expression of the Gamma function.)

\frac{1}{Γ (s)} = s \cdot e^{γ s} \prod_{n = 1}^{\infty} (1 + \frac{s}{n}) e^{- s / n}

(4)

where

γ

is the Euler-Mascheroni constant.

As shown by Riemann,

ζ (s)

extends to

C

as a meromorphic function with only a simple pole at

s = 1

, with residue 1, and satisfies the following functional equation

π^{- \frac{s}{2}} Γ (\frac{s}{2}) ζ (s) = π^{- \frac{1 - s}{2}} Γ (\frac{1 - s}{2}) ζ (1 - s)

(5)

The Riemann zeta function

ζ (s)

has zeros at the negative even integers:

- 2

- 4

- 6

- 8

, ⋯ and one refers to them as the trivial zeros. The other zeros of

ζ (s)

are the complex numbers, i.e., non-trivial zeros [2].

In 1896, Hadamard [4] and Poussin [5] independently proved that no zeros could lie on the line

ℜ (s) = 1

. Together with the functional equation and the fact that there are no zeros with real part greater than 1, this showed that all non-trivial zeros must lie in the interior of the critical strip

0 < ℜ (s) < 1

. This was a key step in their first proofs of the famous Prime Number Theorem.

Later on, Hardy (1914) [6], Hardy and Littlewood (1921) [7] showed that there are infinitely many zeros on the critical line

ℜ (s) = \frac{1}{2}

, which was an astonishing result at that time.

As a summary, we have the following results on the properties of the non-trivial zeros of

ζ (s)

[4,5,6,7,8,9].

Lemma 1.

Non-trivial zeroes of

ζ (s)

, noted as

ρ = α + j β

, have the following properties

1) The number of non-trivial zeroes is infinity;

β \neq 0

;

0 < α < 1

;

ρ, \bar{ρ}, 1 - \bar{ρ}, 1 - ρ

are all non-trivial zeroes.

As further study, a completed zeta function

ξ (s)

is defined as

ξ (s) = \frac{1}{2} s (s - 1) π^{- \frac{s}{2}} Γ (\frac{s}{2}) ζ (s)

(6)

It is well-known that

ξ (s)

is an entire function of order 1. This implies

ξ (s)

is analytic, and can be expressed as infinite polynomial, in the whole complex plane

C

In addition, replacing s with

1 - s

in Equation (6), and combining Equation (5), we have the following functional equation

ξ (s) = ξ (1 - s)

(7)

Considering the definition of

ξ (s)

, and recalling Equation (4), the trivial zeros of

ζ (s)

are canceled by the poles of

Γ (\frac{s}{2})

. The zero of

s - 1

and the pole of

ζ (s)

cancel; the zero

s = 0

and the pole of

Γ (\frac{s}{2})

cancel [9,10]. Thus, all the zeros of

ξ (s)

are exactly the nontrivial zeros of

ζ (s)

. Then we have the following Lemma 2.

Lemma 2.

The zeros of

ξ (s)

coincide with the non-trivial zeros of

ζ (s)

According to Lemma 2, the following two statements for the RH are equivalent.

Statement 1 of the RH: All the non-trivial zeros of

ζ (s)

have real part equal to

\frac{1}{2}

Statement 2 of the RH: All the zeros of

ξ (s)

have real part equal to

\frac{1}{2}

To prove the RH, a natural thinking is to estimate the numbers of non-trivial zeros of

ζ (s)

inside or outside some areas according to Argument Principle. Along this train of thought, there are many research works. Let

N (T)

denote the number of zeros of

ζ (s)

inside the rectangle:

0 < α < 1, 0 < β \leq T

, and let

N_{0} (T)

denote the number of zeros of

ζ (s)

on the line

α = \frac{1}{2}, 0 < β \leq T

. Selberg proved that there exist positive constants c and

T_{0}

, such that

N_{0} (T) > c N (T), (T > T_{0})

[11], later on, Levinson proved that

c \geq \frac{1}{3}

[12], Lou and Yao proved that

c \geq 0.3484

[13], Conrey proved that

c \geq \frac{2}{5}

[14], Bui, Conrey and Young proved that

c \geq 0.41

[15], Feng proved that

c \geq 0.4128

[16], Wu proved that

c \geq 0.4172

[17].

On the other hand, many zeros have been calculated by hand or by computer programs. Among others, Riemann found the first three non-trivial zeros [18]. Gram found the first 15 zeros based on Euler-Maclaurin summation [19]. Titchmarsh calculated the 138^th to 195^th zeros using the Riemann-Siegel formula [20,21]. Here are the first three (pairs of) zeros:

\frac{1}{2} \pm j 14.1347251; \frac{1}{2} \pm j 21.0220396; \frac{1}{2} \pm j 25.0108575

The idea of this paper is originated from Euler’s work on proving the following famous equality

1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + \dots = \frac{π^{2}}{6}

(8)

This interesting result is deduced by comparing the like terms of two types of infinite expressions, i.e., infinite polynomial and infinite product, as shown in the following

\begin{matrix} \frac{sin x}{x} & = 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \dots \\ = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) (1 - \frac{x^{2}}{9 π^{2}}) \dots \end{matrix}

(9)

Then it is conjectured that

ξ (s)

should be factored into

(1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})

or something like that, which was verified by paring

ρ_{i}

and

{\bar{ρ}}_{i}

in the Hadamard product of

ξ (s)

to obtain

(1 - \frac{s}{ρ_{i}}) (1 - \frac{s}{{\bar{ρ}}_{i}}) = \frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})

The Hadamard product of

ξ (s)

as shown in Equation (10) was first proposed by Riemann, however, it was Hadamard [22] who showed the validity of this infinite product expansion.

ξ (s) = ξ (0) \prod_{ρ} (1 - \frac{s}{ρ})

(10)

where

ξ (0) = \frac{1}{2}

ρ

runs over all the non-trivial zeros of the Riemann zeta function

ζ (s)

, or in another word,

ρ

runs over all the zeros of the completed zeta function

ξ (s)

To ensure the absolute convergence of the infinite product expansion,

ρ

and

1 - ρ

are paired. Later in Section 4, we will show that

ρ

and

\bar{ρ}

can also be paired to ensure the absolute convergence of the infinite product expansion.

2. Lemmas

In this section, we first explain the concepts of multiple zeros of

ξ (s)

with their real multiplicities. And then we give three lemmas to support the proof of the RH, in which Lemma 3 is the key lemma.

Multiple zeros of $ξ (s)$ : As shown in Figure 1, the multiple zeros of

ξ (s)

are defined in terms of the quadruplet, i.e.,

ρ, \bar{ρ}, 1 - ρ, 1 - \bar{ρ}

It should be noticed that the multiple zeros with their real multiplicities of

ξ (s)

are objective existence, but the expression of the corresponding factors of

ξ (s)

are optional to some extent. For example, the multiple zeros as shown in Figure 1 have two different expressions as factors of

ξ (s)

and

ξ (1 - s)

, respectively, i.e.,

[(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{2}, (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{2}]

, or

[(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}), (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}), α_{1} + α_{2} = 1, β_{1}^{2} = β_{2}^{2}]

To exclude the latter expression, we stipulate that the multiple zeros

ρ_{i}

related factors of

ξ (s)

take the form of

(1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}}

, where

d_{i} \geq 1

is the real multiplicity of

ρ_{i}

Lemma 3.

Given two infinite products

f (s) = \prod_{i = 1}^{\infty} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}}

(11)

and

f (1 - s) = \prod_{i = 1}^{\infty} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}}

(12)

where s is a complex variable,

ρ_{i} = α_{i} + j β_{i}

and

{\bar{ρ}}_{i} = α_{i} - j β_{i}

are the complex conjugate zeros of

ξ (s)

0 < α_{i} < 1

and

β_{i} \neq 0

are real numbers,

d_{i} \geq 1

are the real multiplicities of

ρ_{i}

, i are natural numbers from 1 to infinity,

β_{i}

are in order of increasing

| β_{i} |

, i.e.,

| β_{1} | \leq | β_{2} | \leq | β_{3} | \leq \dots

Then we have

f (s) = f (1 - s) \Leftrightarrow \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1, 2, 3, \dots, \infty \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots \end{matrix}

(13)

where

" \Leftrightarrow "

is the equivalent sign.

Proof.

First of all, we have the following fact:

\begin{matrix} (1 + \frac{{(s - α)}^{2}}{β^{2}})^{d} = (1 + \frac{{(1 - s - α)}^{2}}{β^{2}})^{d} \\ \Leftrightarrow {(s - α)}^{2} = {(1 - s - α)}^{2} \Leftrightarrow α = \frac{1}{2} \end{matrix}

(14)

where

d \geq 1

is a natural number,

α \neq 0

and

β \neq 0

are real numbers.

Next, the proof is based on Transfinite Induction. Let

P (n)

be:

\begin{matrix} \prod_{i = 1}^{n} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{n} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ \dots \\ (1 + \frac{{(s - α_{n})}^{2}}{β_{n}^{2}})^{d_{n}} = (1 + \frac{{(1 - s - α_{n})}^{2}}{β_{n}^{2}})^{d_{n}} \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1 \dots n \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | \end{matrix} \end{matrix}

(15)

According to Equation (14),

P (1)

is an obvious fact as the Base Case, i.e.,

\begin{matrix} \prod_{i = 1}^{1} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{1} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \Leftrightarrow (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ \Leftrightarrow α_{1} = \frac{1}{2} \end{matrix}

(16)

To be more convincing, let’s further check

P (2)

, i.e.,

\begin{matrix} \prod_{i = 1}^{2} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{2} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} \\ | β_{1} | < | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2} \\ | β_{1} | < | β_{2} | \end{matrix} \end{matrix}

(17)

which is also an obvious fact according to Lemma 5.

As the Successor Case, we need to prove

P (n) \Rightarrow P (n + 1)

Actually, we have

\begin{matrix} \prod_{i = 1}^{n + 1} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{n + 1} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \Leftrightarrow \\ \prod_{i = 1}^{n} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} (1 + \frac{{(s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{d_{n + 1}} = \prod_{i = 1}^{n} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} (1 + \frac{{(1 - s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{d_{n + 1}} \\ \Leftrightarrow (by Lemma 5) \\ \{\begin{matrix} \prod_{i = 1}^{n} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{n} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ (1 + \frac{{(s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{d_{n + 1}} = (1 + \frac{{(1 - s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{d_{n + 1}} \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | < | β_{n + 1} | \end{matrix} \\ \Leftrightarrow (by Equation (15)) \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ \dots \\ (1 + \frac{{(s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{d_{n + 1}} = (1 + \frac{{(1 - s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{d_{n + 1}} \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | < | β_{n + 1} | \end{matrix} \\ \Leftrightarrow (by Equation (14)) \\ \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1, 2, 3, \dots, n, n + 1 \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | < | β_{n + 1} | \end{matrix} \end{matrix}

(18)

Thus the Successor Case is true, i.e.,

P (n) \Rightarrow P (n + 1)

Next, we prove that

P (\infty)

holds by considering well-ordered ordinal set A indexing the family of statements

P (γ : γ \in A)

A = N ⋃ {ω}

with the ordering that

n < ω

for all natural numbers n,

ω

is the first limit ordinal.

It is well-known that

ω = ⋃ {γ : γ < ω}

To prove that

P (\infty)

holds, it suffices to prove the Limit Case, i.e.,

P (γ < ω) \Rightarrow P (ω)

Actually, we have

\begin{matrix} \prod_{i = 1}^{ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \Leftrightarrow \\ \prod_{i = 1}^{γ < ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} (1 + \frac{{(s - α_{ω})}^{2}}{β_{ω}^{2}})^{d_{ω}} = \prod_{i = 1}^{γ < ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} (1 + \frac{{(1 - s - α_{ω})}^{2}}{β_{ω}^{2}})^{d_{ω}} \\ \Leftrightarrow (by Lemma 5) \\ \{\begin{matrix} \prod_{i = 1}^{γ < ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{γ < ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ (1 + \frac{{(s - α_{ω})}^{2}}{β_{ω}^{2}})^{d_{ω}} = (1 + \frac{{(1 - s - α_{ω})}^{2}}{β_{ω}^{2}})^{d_{ω}} \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{ω} | \end{matrix} \\ \Leftrightarrow (by P (α < ω)) \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ \dots \\ (1 + \frac{{(s - α_{ω})}^{2}}{β_{ω}^{2}})^{d_{ω}} = (1 + \frac{{(1 - s - α_{ω})}^{2}}{β_{ω}^{2}})^{d_{ω}} \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{ω} | \end{matrix} \\ \Leftrightarrow (by Equation (14)) \\ \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1, 2, 3, \dots, ω \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{ω} | \end{matrix} \end{matrix}

(19)

Thus the Limit Case is true, i.e.,

P (γ < ω) \Rightarrow P (ω)

Hence we conclude by Transfinite Induction that

P (\infty)

holds, i.e.,

\begin{matrix} \prod_{i = 1}^{\infty} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = \prod_{i = 1}^{\infty} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ \dots \\ (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} = (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}} \\ \dots \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1 \dots \infty \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots \end{matrix} \end{matrix}

(20)

i.e.,

f (s) = f (1 - s) \Leftrightarrow \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1, 2, 3, \dots, \infty \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots \end{matrix}

(21)

That completes the proof of Lemma 3. □

Lemma 4.

Given

(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})

(22)

where s is a complex variable,

0 < α_{i} < 1

and

β_{i} \neq 0

are real numbers,

| β_{1} | \leq | β_{2} |

Then we have

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | \leq | β_{2} | \\ α_{1} + α_{2} = 1, & | β_{1} | = | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}), (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}), & | β_{1} | \leq | β_{2} | \\ (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}), (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}), & | β_{1} | = | β_{2} | \end{matrix} \end{matrix}

(23)

Proof.

Expanding both sides of Equation (22), and comparing the coefficients of like terms, we obtain (details are omitted to save space)

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) \\ \Rightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | \leq | β_{2} | \\ α_{1} + α_{2} = 1, & | β_{1} | = | β_{2} | \end{matrix} \end{matrix}

(24)

The inverse inference of Equation (24) is also an obvious fact. i.e.,

\begin{matrix} \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | \leq | β_{2} | \\ α_{1} + α_{2} = 1, & | β_{1} | = | β_{2} | \end{matrix} \\ \Rightarrow \\ (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) \end{matrix}

Then we have

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | \leq | β_{2} | \\ α_{1} + α_{2} = 1, & | β_{1} | = | β_{2} | \end{matrix} \end{matrix}

(25)

Further, according to Equation (14), i.e.,

(1 + \frac{{(s - α)}^{2}}{β^{2}})^{d} = (1 + \frac{{(1 - s - α)}^{2}}{β^{2}})^{d} \Leftrightarrow α = \frac{1}{2}

and the following similar facts

(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) \Leftrightarrow α_{1} + α_{2} = 1, | β_{1} | = | β_{2} |

(1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) \Leftrightarrow α_{1} + α_{2} = 1, | β_{1} | = | β_{2} |

we have

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | \leq | β_{2} | \\ α_{1} + α_{2} = 1, & | β_{1} | = | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}), (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}), & | β_{1} | \leq | β_{2} | \\ (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}), (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}), & | β_{1} | = | β_{2} | \end{matrix} \end{matrix}

(26)

That completes the proof of Lemma 4. □

Remark 1.

From the viewpoint of the products of (polynomial) factors, Lemma 4 reveals a fact that there are only two possible cases to make Eq. (22) hold, i.e., Case 1:

(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})

(1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})

; and Case 2:

(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}) = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})

(1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}) = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})

, respectively. This result is quite different from the similar equation of products of factors

a \cdot b = c \cdot d

, which has infinite cases of solutions for real numbers

a, b, c, d

Remark 2.

Given

(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}

(27)

where s is a complex variable,

0 < α_{1} < 1, 0 < α_{2} < 1

and

β_{1} \neq 0, β_{2} \neq 0

are real numbers,

d_{1} \geq 1, d_{2} \geq 1

are natural numbers, denoting the real multiplicities of

ρ_{1} = α_{1} + j β_{1}

and

ρ_{2} = α_{2} + j β_{2}

, respectively,

| β_{1} | \leq | β_{2} |

Then we have

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2} \\ | β_{1} | < | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} \\ | β_{1} | < | β_{2} | \end{matrix} \end{matrix}

(28)

Proof.

Based on Lemma 4, and considering additional possibilities that

(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} ∣ (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}

(1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} ∣ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}

(where "∣" is the divisible sign), or vice versa, we have

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}}, (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}, & | β_{1} | < | β_{2} | \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} ∣ (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}, (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} ∣ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} \\ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2} - d_{1}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2} - d_{1}} \end{matrix}, & | β_{1} | = | β_{2} |, d_{1} < d_{2} \\ \{\begin{matrix} (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} ∣ (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}}, (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} ∣ (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1} - d_{2}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1} - d_{2}} \end{matrix}, & | β_{1} | = | β_{2} |, d_{1} > d_{2} \\ (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}, (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}}, & | β_{1} | = | β_{2} |, d_{1} = d_{2} \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | < | β_{2} | \\ α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | = | β_{2} |, d_{1} < d_{2} ★ \\ α_{1} = α_{2} = \frac{1}{2}, & | β_{1} | = | β_{2} |, d_{1} > d_{2} ★ \\ α_{1} + α_{2} = 1, & | β_{1} | = | β_{2} |, d_{1} = d_{2} ★ \end{matrix} \\ \Leftrightarrow \\ (To exclude the duplication situations marked by ★ between the first quadruplet of zeros represented \\ by ρ_{1} = α_{1} + j β_{1} \\ and the \sec ond quadruplet of zeros represented by ρ_{2} = α_{2} + j β_{2}, which contradict the given condition \\ that the real multiplicities of ρ_{1} and ρ_{2} are d_{1} and d_{2}, respectively .) \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2} \\ | β_{1} | < | β_{2} | \end{matrix} \\ \Leftrightarrow (by Equation (14)) \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{d_{1}} \\ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{d_{2}} \\ | β_{1} | < | β_{2} | \end{matrix} \end{matrix}

(29)

That completes the proof of Lemma 5. □

3. A Proof of the RH

This section is planned to present a proof of the Riemann Hypothesis. We first prove that Statement 2 of the RH is true, and then by Lemma 2, Statement 1 of the RH is also true.

Proof of the RH.

The details are delivered in three steps as follows.

Step 1: It is well-known that all the zeros of

ξ (s)

always come in complex conjugate pairs. Then by pairing

ρ_{i} = α_{i} + j β_{i}

and

{\bar{ρ}}_{i} = α_{i} - j β_{i}

in the Hadamard product as shown in Equation (10), we have

\begin{matrix} ξ (s) & = ξ (0) \prod_{ρ} (1 - \frac{s}{ρ}) \\ = ξ (0) \prod_{i = 1}^{\infty} (1 - \frac{s}{ρ_{i}}) (1 - \frac{s}{{\bar{ρ}}_{i}}) \\ = ξ (0) \prod_{i = 1}^{\infty} (1 - \frac{s}{α_{i} + j β_{i}}) (1 - \frac{s}{α_{i} - j β_{i}}) \\ = ξ (0) \prod_{i = 1}^{\infty} (\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}}) \end{matrix}

(30)

where

ξ (0) = \frac{1}{2}, 0 < α_{i} < 1, β_{i} \neq 0

The absolute convergence of the infinite product in Equation (30) in the form

ξ (s) = ξ (0) \prod_{i = 1}^{\infty} (1 - \frac{s}{ρ_{i}}) (1 - \frac{s}{{\bar{ρ}}_{i}}) = ξ (0) \prod_{i = 1}^{\infty} (1 - \frac{s (2 α_{i} - s)}{| ρ_{i} |^{2}})

(31)

depends on the convergence of infinite series

\sum_{i = 1}^{\infty} \frac{1}{| ρ_{i} |^{2}}

, or equivalently,

\sum_{ρ} \frac{1}{{| ρ |}^{2}}

, which is an obvious fact according to Theorem 2 in Section 2, Chapter IV of Ref. [23], as shown in the following.

Theorem 1.

([23]).The function

ξ (s)

is an entire function of order one that has infinitely many zeros

ρ_{n}

such that

0 \leq Re ρ_{n} \leq 1

. The series

\sum | ρ_{n} |^{- 1}

diverges, but the series

\sum | ρ_{n} |^{- 1 - ε}

converges for any

ε > 0

. The zeros of

ξ (s)

are the nontrivial zeros of

ζ (s)

Remark 3.

In the Theorem 2 of Ref. [23],

Re (\cdot)

is identical to

ℜ (\cdot)

in this paper, both

Re (\cdot)

and

ℜ (\cdot)

mean the real part of any complex number.

Further, considering the absolute convergence of

ξ (s) = ξ (0) \prod_{i = 1}^{\infty} (1 - \frac{s (2 α_{i} - s)}{| ρ_{i} |^{2}}) = ξ (0) \prod_{i = 1}^{\infty} (\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}})

(32)

we have the following new expression of

ξ (s)

by putting all the

ρ_{i}

related multiple factors (zeros) together in the above Equation (32)

ξ (s) = ξ (0) \prod_{i = 1}^{\infty} (\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}})^{d_{i}}

(33)

where

d_{i} \geq 1

are the real multiplicities of

ρ_{i}

, i are natural numbers from 1 to infinity.

Step 2: Replacing s with

1 - s

in Equation (33), we obtain the infinite product expression of

ξ (1 - s)

, i.e.,

ξ (1 - s) = ξ (0) \prod_{i = 1}^{\infty} {(\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(1 - s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}})}^{d_{i}}

(34)

Remark 4.

According to the new expressions of

ξ (s)

and

ξ (1 - s)

, i.e., Equation (33) and Equation (34), we may conclude that all

ρ_{i}

and

1 - ρ_{i}

related multiple zeros, i.e.,

{(α_{i} \pm j β_{i})}^{d_{i}}, {(1 - α_{i} \pm j β_{i})}^{d_{i}}

are included in the

i^{t h}

group of factors,

(1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}}

and

(1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{d_{i}}

, respectively, or in another word, before or after the

i^{t h}

group of factors of

ξ (s)

and

ξ (1 - s)

, there are no

ρ_{i}

and

1 - ρ_{i}

related multiple zeros.

Actually, with such arrangement of $ρ_{i}$ and $1 - ρ_{i}$ related multiple factors of $ξ (s)$ and $ξ (1 - s)$ , we "assigned" a reason for excluding, in the proof of Lemma 5 and Lemma 3, the "abnormal" situation, i.e., the successor factor and its predecessor factor represent the same quadruplet of zeros.

Step 3: According to the functional equation

ξ (s) = ξ (1 - s)

, and considering Equation (33) and Equation (34), we have

ξ (0) \prod_{i = 1}^{\infty} {(\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}})}^{d_{i}} = ξ (0) \prod_{i = 1}^{\infty} {(\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(1 - s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}})}^{d_{i}}

(35)

which is equivalent to

\prod_{i = 1}^{\infty} {(1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})}^{d_{i}} = \prod_{i = 1}^{\infty} {(1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})}^{d_{i}}

(36)

And that

β_{i}

can be certainly arranged in order of increasing

| β_{i} |

, i.e.,

| β_{1} | \leq | β_{2} | \leq | β_{3} | \leq \dots

Then, according to Lemma 3, Equation (36) is equivalent to

\{\begin{matrix} α_{i} = \frac{1}{2}, i = 1, 2, 3, \dots, \infty \\ | β_{1} | < | β_{2} | < | β_{3} | < \dots \end{matrix}

Thus, we conclude that all the zeros of the completed zeta function

ξ (s)

have real part equal to

\frac{1}{2}

, i.e., Statement 2 of the RH is true. According to Lemma 2, Statement 1 of the RH is also true, i.e., all the non-trivial zeros of the Riemann zeta function

ζ (s)

have real part equal to

\frac{1}{2}

That completes the proof of the RH. □

Remark 5.

By Lemma 1, there are 2 pairs of complex zeros of

ζ (s)

simultaneously, i.e.,

ρ = α + j β, \bar{ρ} = α - j β, 1 - ρ = 1 - α - j β, 1 - \bar{ρ} = 1 - α + j β

are all the non-trivial zeroes of

ζ (s)

. With the proof of the RH, i.e.,

α = \frac{1}{2}

, these 2 pairs of zeros are actually only one pair, because

ρ = 1 - \bar{ρ} = \frac{1}{2} + j β, \bar{ρ} = 1 - ρ = \frac{1}{2} - j β

. Thus Lemma 1 could be modified more precisely as follows.

Lemma 5.

Non-trivial zeroes of

ζ (s)

, noted as

ρ = α + j β

, have the following properties

1) The number of non-trivial zeroes is infinity;

β \neq 0

;

0 < α < 1

;

ρ = 1 - \bar{ρ}, \bar{ρ} = 1 - ρ

are all non-trivial zeroes.

4. Conclusion

The celebrated Riemann Hypothesis is proved to be true based on a new expression of the completed zeta function

ξ (s)

, i.e.,

ξ (s) = ξ (0) \prod_{i = 1}^{\infty} (\frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} + \frac{{(s - α_{i})}^{2}}{α_{i}^{2} + β_{i}^{2}})^{d_{i}}

where

ξ (0) = \frac{1}{2}, ρ_{i} = α_{i} + j β_{i}

and

{\bar{ρ}}_{i} = α_{i} - j β_{i}

are the complex conjugate zeros of

ξ (s)

0 < α_{i} < 1

and

β_{i} \neq 0

are real numbers,

d_{i} \geq 1

are the real multiplicities of

ρ_{i}

, i are natural numbers from 1 to infinity,

β_{i}

are in order of increasing

| β_{i} |

, i.e.,

| β_{1} | \leq | β_{2} | \leq | β_{3} | \leq \dots

Acknowledgments

The author would like to gratefully acknowledge the help received from Prof. Tianguang Chu (Peking University) while preparing this article.

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Figure 1. Illustration of the multiple zeros of

ξ (s)

Figure 1. Illustration of the multiple zeros of

ξ (s)

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A Proof Of The Riemann Hypothesis Based On A New Expression Of The Completed Zeta Function

Abstract

1. Introduction

2. Lemmas

3. A Proof of the RH

4. Conclusion

Acknowledgments

References

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