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:

11 June 2024

Posted:

14 June 2024

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Abstract

The Riemann Hypothesis (RH) is proved based on a new absolutely convergent expression of $\xi(s)$, which was obtained from the Hadamard product, through paring $\rho_i$ and $\bar{\rho}_i$, and taking the possible multiple zeros into consideration with their real (unique and unchangeable) multiplicities, i.e. $$\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho})=\xi(0)\prod_{i=1}^{\infty}(1-\frac{s}{\rho_i})(1-\frac{s}{\bar{\rho}_i})=\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}}$$ 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, $d_i\geq 1$ is the real multiplicity of $\rho_i$, $\beta_i$ are arranged in order of increasing $|\beta_i|$, i.e., $0<|\beta_1|\leq|\beta_2|\leq|\beta_3|\leq \cdots$, $i =1,2,3, \cdots, \infty$.\\ Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\prod_{i=1}^{\infty}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)}^{d_{i}}=\prod_{i=1}^{\infty}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}^{d_{i}}$$ which, owing to the uniqueness and unchangeableness of $d_i$ (see Lemma 3 for the proof details), is equivalent to $$\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)}^{d_{i}}=\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}^{d_{i}} \Leftrightarrow \alpha_i=\frac{1}{2}, 0<|\beta_1|<|\beta_2|<|\beta_3|<\cdots; i =1,2,3, \cdots, \infty$$ Thus, we conclude that the RH is true.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

The Riemann Hypothesis

^{[1]}

is one of the most important unsolved problems in mathematics. Although many efforts and achievements have been made towards proving this celebrated hypothesis, it still remains an open problem

^{[2 - 3]}

. The Riemann zeta function is originally 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 above-defined Riemann zeta function and prime numbers was discovered by Euler, i.e., the famous Euler product

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 in his paper in 1859 how to extend the zeta function to the whole complex plane

C

by analytic continuation, i.e.

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

(3a)

where

" \int_{\infty}^{\infty} "

is the symbol adopted by Riemann to represent the contour integral from

+ \infty

+ \infty

around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior.

Or equivalently,

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}

(3b)

where

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

is the Jaccobi theta function,

Γ

is the Gamma function in the following Weierstrass expression

\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

ξ (s) = ξ (1 - s)

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

. 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}

To give a summary of the related research on the RH, we have the following results on the properties of the non-trivial zeros of

ζ (s)

^{[4 - 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;

(2)

β \neq 0

;

(3)

0 < α < 1

;

(4)

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 obtain 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)

Accordingly, the following two statements of the RH are equivalent.

Statement 1: All the non-trivial zeros of

ζ (s)

have real part equal to

\frac{1}{2}

Statement 2: 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 certain areas according to Argument Principle. Along this train of thought, there are many research works. Let

N (T)

denote the number of non-trivial zeros of

ζ (s)

inside the rectangle:

0 < α < 1, 0 < β \leq T

, and let

N_{0} (T)

denote the number of non-trivial 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 non-trivial 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) non-trivial 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 the author of this paper 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)

, i.e.

(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 who showed the validity of this infinite product expansion

^{[22]}

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

(10)

where

ξ (0) = \frac{1}{2}

ρ

runs over all the zeros of the completed zeta function

ξ (s)

Hadamard pointed out that to ensure the absolute convergence of the infinite product expansion,

ρ

and

1 - ρ

are paired. Later in Section 3, 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 concept of the real multiplicity of a zero of

ξ (s)

. And then we give four lemmas (Lemma 3 to Lemma 6) to support the proof of the RH, among which Lemma 3 is the key lemma.

Multiple zeros of $ξ (s)$ and their real multiplicities: 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}

Figure 1. Illustration of the multiple zeros of

ξ (s)

Figure 1. Illustration of the multiple zeros of

ξ (s)

There are two different expressions of factors of

ξ (s) / ξ (1 - s)

for the multiple zeros in Figure 1, 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}})

with

α_{1} + α_{2} = 1, β_{1}^{2} = β_{2}^{2}

To exclude the latter expression, we stipulate that zero

ρ_{i}

related factors of

ξ (s) / ξ (1 - s)

take the unique form of

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

, where

m_{i} \geq 1

is the real multiplicity of

ρ_{i}

, here "real" means unique and unchangeable. In Figure 1, the multiplicity of

ρ_{1}

is 2, i.e.,

m_{1} = 2

Remark: Although the real multiplicity

m_{i}

of zero

ρ_{i}

is unknown, it is an objective existence, unique, and unchangeable. This is the key point in the proofs of Lemma 3, Lemma 4, and Lemma 5.

Lemma 3: Given two absolutely convergent infinite products

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

(11)

and

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

m_{i} \geq 1

is the real multiplicity of

ρ_{i}

0 < | 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} \\ 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots \\ i = 1, 2, 3, \dots, \infty \end{matrix}

(13)

where

" \Leftrightarrow "

is the equivalent sign.

Proof: First of all, we have the following fact:

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

(14)

where

m \geq 1

is an integer,

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}})^{m_{i}} = \prod_{i = 1}^{n} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} \\ \dots \\ (1 + \frac{{(s - α_{n})}^{2}}{β_{n}^{2}})^{m_{n}} = (1 + \frac{{(1 - s - α_{n})}^{2}}{β_{n}^{2}})^{m_{n}} \\ 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{i} = \frac{1}{2}, i = 1 \dots n \\ 0 < | 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}})^{m_{i}} = \prod_{i = 1}^{1} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \\ \Leftrightarrow (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} \Leftrightarrow α_{1} = \frac{1}{2} \end{matrix}

(16)

To be more convincing, let’s further check

P (2)

, i.e.,

(17)

which is also an obvious fact according to Lemma 4.

As the Successor Case, we need to prove

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

Actually, we have

(17)

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)

Based on Equation(14), it is an obvious fact that

α_{i} = \frac{1}{2} \Rightarrow \prod_{i = 1}^{ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}}

Then we only need to prove, under condition

P (γ < ω)

, that

\prod_{i = 1}^{ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \Rightarrow α_{i} = \frac{1}{2}

(19)

For the sake of contradiction, assume

\prod_{i = 1}^{ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} ⇏ α_{i} = \frac{1}{2}

(19a)

under condition

P (γ < ω)

, i.e.

\prod_{i = 1}^{γ < ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{γ < ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \Leftrightarrow α_{i} = \frac{1}{2},

(19b)

then we have

\prod_{i = 1}^{ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \Rightarrow α_{i} \neq \frac{1}{2}

(19c)

[Otherwise, since

f (s) / f (1 - s)

are absolutely convergent, then through rearrangement of the order of factors in the left-hand part of Equation(19c), we have

α_{i} \neq \frac{1}{2}, i = 1, . . ., γ, γ < ω

, and

α_{i} = \frac{1}{2}, i = γ + 1, . . ., ω

(at least

i = ω

). Further, based on Equation(14), canceling the corresponding factors of these

α_{i} = \frac{1}{2}

in the left-hand part of Equation(19c), we obtain

\prod_{i = 1}^{γ < ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{γ < ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}}

(19d)

Taking Equation(19b) into consideration, Equation(19d) leads to

α_{i} = \frac{1}{2}, i = 1, . . ., γ, γ < ω

, this together with

α_{i} = \frac{1}{2}, i = γ + 1, . . ., ω

, is a contradiction to the assumption Equation(19a).]

Next, we shall show that Equation(19c) contradicts the "divisibility" implied in

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

We have

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

(19e)

It follows from Equation(19e) that

(1 + \frac{{(1 - s - α_{ω})}^{2}}{β_{ω}^{2}})^{m_{ω}} | \prod_{i = 1}^{γ < ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} (1 + \frac{{(s - α_{ω})}^{2}}{β_{ω}^{2}})^{m_{ω}}

(19f)

where

" | "

is the "divisible" sign.

According to Lemma 6, Equation(19f) implies that

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

either divides

\prod_{i = 1}^{γ < ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}}

(this leads to the violation of the real multiplicities of zeros, similar to the situation in the proofs of Lemma 4 and Lemma 5) or divides

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

(this is impossible, because

α_{ω} \neq \frac{1}{2}

). Then we come to an absurdity that

\prod_{i = 1}^{ω} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \neq \prod_{i = 1}^{ω} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}}

Thus, assumption Equation(19a) does not hold under condition

P (γ < ω)

Consequently, Equation(19) holds under condition

P (γ < ω)

To keep the real multiplicities of zeros unchanged, the imaginary parts

β_{i}

should be further limited to

0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{ω} |

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}})^{m_{i}} = \prod_{i = 1}^{\infty} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \\ 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots \\ i = 1, 2, 3, \dots, \infty \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{i} = \frac{1}{2} \\ 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots \\ i = 1, 2, 3, \dots, \infty \end{matrix} \end{matrix}

(20)

i.e.,

f (s) = f (1 - s) \Leftrightarrow \{\begin{matrix} α_{i} = \frac{1}{2} \\ 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots \\ i = 1, 2, 3, \dots, \infty \end{matrix}

(21)

That completes the proof of Lemma 3.

Lemma 4: Given

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

(22)

where s is a complex variable,

0 < α_{1} < 1, 0 < α_{2} < 1, β_{1} \neq 0, β_{2} \neq 0

are real numbers,

0 < | β_{1} | \leq | β_{2} |

m_{1} \geq 1

and

m_{2} \geq 1

are the real multiplicities of

ρ_{1} = α_{1} + j β_{1}

and

ρ_{2} = α_{2} + j β_{2}

, respectively.

Then we have

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{m_{2}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{m_{2}} \\ \Leftrightarrow \\ \{\begin{matrix} {(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})}^{m_{1}} = {(1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})}^{m_{1}} \\ {(1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})}^{m_{2}} = {(1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})}^{m_{2}} \\ 0 < | β_{1} | < | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2} \\ 0 < | β_{1} | < | β_{2} | \end{matrix} \end{matrix}

(23)

Proof: Equation(22) has an obvious solution, i.e.,

\begin{matrix} \{\begin{matrix} {(1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})}^{m_{1}} = {(1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})}^{m_{1}} \\ {(1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})}^{m_{2}} = {(1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})}^{m_{2}} \\ 0 < | β_{1} | \leq | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2} \\ 0 < | β_{1} | \leq | β_{2} | \end{matrix} \end{matrix}

(24)

Next, we shall show that the above obvious solution is the only solution to Equation(22). Then let us try to find other solutions of Equation(22) by considering

α_{1} \neq \frac{1}{2}, α_{2} \neq \frac{1}{2} \Leftrightarrow g c d (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}}, 1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}}) = g c d (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}}, 1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}}) = 1

(25)

where "gcd" stands for: greatest common divisor.

Based on Equation(22), and suppose

m_{2} \geq m_{1}

without loss of generality, we have

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

(26)

where

" | "

is the "divisible" sign.

By Lemma 6 and Equation(25), it follows that

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

(27)

Then we have

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

(28)

By comparing like terms of both sides of Equation(28), it follows that

α_{1} + α_{2} = 1, | β_{1} | = | β_{2} |

. Consequently,

α_{2} \pm j β_{2} / 1 - α_{2} \pm j β_{2}

is a duplicate of

α_{1} \pm j β_{1} / 1 - α_{1} \pm j β_{1}

in terms of quadruplet, which is the same situation as shown in Figure 1. That contradicts the definition of real multiplicities in Equation(22), i.e., in Equation(22) the real multiplicity of

α_{1} \pm j β_{1} / 1 - α_{1} \pm j β_{1}

m_{1}

, if

α_{1} + α_{2} = 1, | β_{1} | = | β_{2} |

, then the real multiplicity of

α_{1} \pm j β_{1} / 1 - α_{1} \pm j β_{1}

would be

m_{1} + m_{2}

Therefore, the obvious solution Equation(24) is the only solution of Equation(22), i.e.

\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{m_{2}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{m_{2}} \\ \Leftrightarrow \\ \{\begin{matrix} (1 + \frac{{(s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} = (1 + \frac{{(1 - s - α_{1})}^{2}}{β_{1}^{2}})^{m_{1}} \\ (1 + \frac{{(s - α_{2})}^{2}}{β_{2}^{2}})^{m_{2}} = (1 + \frac{{(1 - s - α_{2})}^{2}}{β_{2}^{2}})^{m_{2}} \\ 0 < | β_{1} | \leq | β_{2} | \end{matrix} \\ \Leftrightarrow \\ \{\begin{matrix} α_{1} = α_{2} = \frac{1}{2} \\ 0 < | β_{1} | \leq | β_{2} | \end{matrix} \end{matrix}

(29)

Further excluding the situation that violates the definition of the real multiplicities of zeros in Equation(29), i.e.,

α_{1} = α_{2} = \frac{1}{2}, | β_{1} | = | β_{2} |

, then we know that Equation(23) holds.

That completes the proof of Lemma 4.

Another Proof of Lemma 4: It is obvious that each factor in Equation(22) may divide other two factors on the opposite side. Consequently, Equation(22) can be rewritten as follows

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

(30)

or (suppose

m_{2} \geq m_{1}

without loss of generality)

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

(31)

k and

k^{'}

are constants, which can be easily determined as

k = k^{'} = 1

by comparing the like terms of related polynomial equations. Then, Equation(30) is equivalent to

α_{1} = α_{2} = \frac{1}{2}

, but Equation(31) yields solution:

α_{1} + α_{2} = 1, | β_{1} | = | β_{2} |

, which contradicts the definition of the real multiplicities of zeros. After discarding Equation(31), and further excluding the potential situation

α_{1} = α_{2} = 1, | β_{1} | = | β_{2} |

in the solution of Equation(30), we know that Equation(23) holds.

That completes the proof of Lemma 4.

Lemma 5: Given

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

(32)

where s is a complex variable,

0 < α_{i} < 1

and

β_{i} \neq 0

are real numbers,

0 < | β_{1} | \leq | β_{2} | \leq | β_{3} | \dots \leq | β_{n + 1} |

m_{i} \geq 1

is the real multiplicity of

ρ_{i} = α_{i} + j β_{i}

Then we have

\begin{matrix} \prod_{i = 1}^{n + 1} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{n + 1} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \\ \Leftrightarrow \\ \{\begin{matrix} \prod_{i = 1}^{n} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} = \prod_{i = 1}^{n} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}} \\ (1 + \frac{{(s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{m_{n + 1}} = (1 + \frac{{(1 - s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{m_{n + 1}} \\ 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots < | β_{n} | < | β_{n + 1} | \end{matrix} \end{matrix}

(33)

Proof: Equation(32) is equivalent to

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

(34)

Similar to "Another proof of Lemma 4", Equation(34) can be rewritten as follows

\frac{\prod_{i = 1}^{n} (1 + \frac{{(s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}}}{\prod_{i = 1}^{n} (1 + \frac{{(1 - s - α_{i})}^{2}}{β_{i}^{2}})^{m_{i}}} = \frac{(1 + \frac{{(1 - s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{m_{n + 1}}}{(1 + \frac{{(s - α_{n + 1})}^{2}}{β_{n + 1}^{2}})^{m_{n + 1}}} = k = 1

(35)

or (suppose

\sum_{i = 1}^{n} m_{i} \geq m_{n + 1}

without loss of generality)

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

(36)

Then after Equation(36) is abandoned (for that will generate solutions violating the definition of the real multiplicities of zeros) and further excluding the potential situations:

α_{i} = α_{l} = \frac{1}{2}, | β_{i} | = | β_{l} |, i \neq l, 1 \leq i \leq n + 1, 1 \leq l \leq n + 1

, in the solution of Equation(35), we know that Equation(33) holds.

That completes the proof of Lemma 5.

Lemma 6: Let F be a field,

p (x), q (x), m (x) \in F [x]

. If

m (x)

divides the product

p (x) q (x)

, but

m (x)

and

p (x)

are relatively prime, i.e.,

g c d (m (x), p (x)) = 1

, then

m (x)

divides

q (x)

, where

F [x]

is defined as the set of all polynomials in x over F:

F [x] = {\sum_{i = 0}^{\infty} a_{i} x^{i} | a_{i} \in F, a_{i} = 0 for all but a finite number of i}

Remark: The set

F [x]

equipped with the operations + and · is the polynomial ring in x over the field F.

Remark: The content of Lemma 6 (Polynomial Algebra over Fields) can be found in many textbooks of Linear Algebra or Advanced Algebra. It is the foundation to find other possible solutions in addition to the obvious solution in Lemma 4 and 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. To be brief, to prove the Riemann Hypothesis, it suffices to show that

α_{i} = \frac{1}{2}, i = 1, 2, 3, \dots, \infty

in the new expression of

ξ (s)

as shown in Equation(37).

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}

(37)

where

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

The absolute convergence of the infinite product in Equation(37) 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}})

(38)

depends on the convergence of infinite series

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

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

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}})

(39)

we have the following new expression of

ξ (s)

by putting all the

ρ_{i}

related multiple factors (zeros) together in the above Equation(39)

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}})^{m_{i}}

(40)

where

m_{i} \geq 1

is the real multiplicity of

ρ_{i}

i = 1, 2, 3, \dots, \infty

Step 2: Replacing s with

1 - s

in Equation(40), 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}})}^{m_{i}}

(41)

Step 3: According to the functional equation

ξ (s) = ξ (1 - s)

, and considering Equation(40) and Equation(41), 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}})}^{m_{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}})}^{m_{i}}

(42)

which is equivalent to

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

(43)

where

β_{i}

are in order of increasing

| β_{i} |

, i.e.,

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

To check the absolute convergence of both sides of Equation(43), it suffices to make a comparison with Equation(38) without considering multiple zeros in Equation (43), i.e., to make a comparison between

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

and

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

. It is well-known that the absolute convergence of

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

depends on the convergence of infinite series

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

(already proved in Step 1); the absolute convergence of

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

depends on the convergence of infinite series

\sum_{i = 1}^{\infty} \frac{1}{β_{i}^{2}}

, which is also an obvious fact because

0 < α_{i} < 1, | ρ_{i} | \to \infty, | β_{i} | \to \infty, as i \to \infty, {lim}_{i \to \infty} \frac{β_{i}^{2}}{| ρ_{i} |^{2}} = {lim}_{i \to \infty} \frac{β_{i}^{2}}{α_{i}^{2} + β_{i}^{2}} = 1

, that means

\sum_{i = 1}^{\infty} \frac{1}{β_{i}^{2}}

and

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

have the same convergence.

Then, according to Lemma 3, Equation(43) is equivalent to

α_{i} = \frac{1}{2}; 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots; i = 1, 2, 3, \dots, \infty

(44)

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.

4. Retrospection and Discussion

On the Simultaneous Zeros of $ξ (s)$

According to Lemma 1, there are two 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

. With the proof of the RH, 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 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;

(2)

β \neq 0

;

(3)

0 < α < 1

;

(4)

ρ = 1 - \bar{ρ}, \bar{ρ} = 1 - ρ

are all non-trivial zeroes.

On the Paring of Zeros of $ξ (s)$

Hadamard pointed out that to ensure the absolute convergence of the Hadamard product, i.e.,

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

ρ

and

1 - ρ

are paired. In Section 3, the author proved that

ρ

and

\bar{ρ}

can also be paired to ensure the absolute convergence of the Hadamard product. And that the paring of conjugate zeros, i.e.,

ρ

and

\bar{ρ}

, is the right way to express the most essential characteristic of

ξ (s)

as (infinite) polynomial with real coefficients, whereas

1 - ρ

and

1 - \bar{ρ}

are just another pair of conjugate zeros given by

ξ (s) = ξ (1 - s)

5. Conclusion

This paper presents a proof of the RH based on a new expression of

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}})^{m_{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,

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

m_{i} \geq 1

is the real multiplicity of

ρ_{i}

The proof is conducted according to Transfinite Induction. The first key-point is the paring of conjugate zeros

ρ

and

\bar{ρ}

to get the new expression of

ξ (s)

. The second key-point is the use of "real multiplicities" of the zeros of

ξ (s)

. Obviously, the real multiplicity of a zero of

ξ (s)

is an objective existence, unique, and unchangeable. As a result, the functional equation

ξ (s) = ξ (1 - s)

finally leads to

α_{i} = \frac{1}{2}; 0 < | β_{1} | < | β_{2} | < | β_{3} | < \dots; i = 1, 2, 3, \dots, \infty

Acknowledgments

The author would like to gratefully acknowledge the help received from Dr. Victor Ignatov (Independent Researcher), Prof. Mark Kisin (Harvard University), Prof. Yingmin Jia (Beihang University), Prof. Tianguang Chu (Peking University), Prof. Guangda Hu (Shanghai University), Prof. Jiwei Liu (University of Science and Technology Beijing), and Dr. Shangwu Wang (Beijing 99view Technology Limited), while preparing this article. The author is also grateful to the editors and referees of PNAS for their constructive comments and suggestions. Finally, with this manuscript, the author pays tribute to Bernhard Riemann and other predecessor mathematicians. They are the shining stars in the sky of human civilization. This manuscript has no associated data.

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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. Retrospection and Discussion

5. Conclusion

Acknowledgments

References

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