A Proof of the Riemann Hypothesis Based on a New Expression of the Completed Zeta Function

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 $\Re \left(s\right)>1$ by the absolutely convergent series ^[2]

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}},\Re \left(s\right)>1$$

(1)

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

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}}=\prod _p{(1-p^{-s})}^{-1},\Re \left(s\right)>1$$

(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 $\mathbb{C}$ by analytic continuation, i.e.,

$$\zeta \left(s\right)={\displaystyle \frac{\Gamma (1-s)}{2\pi i}}{\int }_{\infty }^{\infty }{\displaystyle \frac{{(-x)}^s}{e^x-1}}·{\displaystyle \frac{dx}{x}}$$

(3a)

where $"{\int }_{\infty }^{\infty }"$ is the symbol adopted by Riemann to represent the contour integral from $+\infty$ to $+\infty$ around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior.

Or equivalently,

$$\zeta \left(s\right)={\displaystyle \frac{{\pi }^{s/2}}{\Gamma (s/2)}}\{{\displaystyle \frac{1}{s(s-1)}}+{\int }_1^{\infty }(x^{{\displaystyle \frac{s}{2}}-1}+x^{-{\displaystyle \frac{s}{2}}-{\displaystyle \frac{1}{2}}})·\left({\displaystyle \frac{\theta \left(x\right)-1}{2}}\right)dx\}$$

(3b)

where $\theta \left(x\right)={\sum }_{-\infty }^{\infty }{e}^{-n^2\pi x}$ is the Jaccobi theta function, $\Gamma$ is the Gamma function in the following Weierstrass expression

$${\displaystyle \frac{1}{\Gamma \left(s\right)}}=s·e^{\gamma s}\prod _{n=1}^{\infty }(1+{\displaystyle \frac{s}{n}})e^{-s/n}$$

(4)

where $\gamma$ is the Euler-Mascheroni constant.

As shown by Riemann, $\zeta \left(s\right)$ extends to $\mathbb{C}$ as a meromorphic function with only a simple pole at $s=1$, with residue 1, and satisfies the following functional equation

$${\pi }^{-{\displaystyle \frac{s}{2}}}\Gamma \left({\displaystyle \frac{s}{2}}\right)\zeta \left(s\right)={\pi }^{-{\displaystyle \frac{1-s}{2}}}\Gamma \left({\displaystyle \frac{1-s}{2}}\right)\zeta (1-s)$$

(5)

The Riemann zeta function $\zeta \left(s\right)$ has zeros at the negative even integers: $-2$, $-4$, $-6$, $-8$, ⋯ and one refers to them as the trivial zeros. The other zeros of $\zeta \left(s\right)$ 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 $\Re \left(s\right)=1$, together with the functional equation $\xi \left(s\right)=\xi (1-s)$ 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$0<\Re \left(s\right)<1$. Later on, Hardy (1914) ^[6], Hardy and Littlewood (1921) ^[7] showed that there are infinitely many zeros on the critical line$\Re \left(s\right)={\displaystyle \frac{1}{2}}$.

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 $\zeta \left(s\right)$  ^[4–9].

Lemma 1: Non-trivial zeroes of $\zeta \left(s\right)$, noted as $\rho =\alpha +j\beta$, have the following properties

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

2) $\beta \ne 0$;

3) $0<\alpha <1$;

4) $\rho ,\overline{\rho },1-\overline{\rho },1-\rho$ are all non-trivial zeroes.

As further study, a completed zeta function $\xi \left(s\right)$ is defined as

$$\xi \left(s\right)={\displaystyle \frac{1}{2}}s(s-1){\pi }^{-{\displaystyle \frac{s}{2}}}\Gamma \left({\displaystyle \frac{s}{2}}\right)\zeta \left(s\right)$$

(6)

It is well-known that $\xi \left(s\right)$ is an entire function of order 1. This implies $\xi \left(s\right)$ is analytic, and can be expressed as infinite polynomial, in the whole complex plane $\mathbb{C}$. In addition, replacing s with $1-s$ in Equation (6), and combining Equation (5), we obtain the following functional equation

$$\xi \left(s\right)=\xi (1-s)$$

(7)

Considering the definition of $\xi \left(s\right)$, and recalling Equation (4), the trivial zeros of $\zeta \left(s\right)$ are canceled by the poles of $\Gamma \left({\displaystyle \frac{s}{2}}\right)$. The zero of $s-1$ and the pole of $\zeta \left(s\right)$ cancel; the zero $s=0$ and the pole of $\Gamma \left({\displaystyle \frac{s}{2}}\right)$ cancel ^[9–10]. Thus, all the zeros of $\xi \left(s\right)$ are exactly the nontrivial zeros of $\zeta \left(s\right)$. Then we have the following Lemma 2.

Lemma 2: The zeros of $\xi \left(s\right)$ coincide with the non-trivial zeros of $\zeta \left(s\right)$.

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

Statement 1: All the non-trivial zeros of $\zeta \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

Statement 2: All zeros of $\xi \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

To prove the RH, a natural thinking is to estimate the numbers of non-trivial zeros of $\zeta \left(s\right)$ inside or outside some certain areas according to Argument Principle. Along this train of thought, there are many research works. Let $N\left(T\right)$ denote the number of non-trivial zeros of $\zeta \left(s\right)$ inside the rectangle: $0<\alpha <1,0<\beta \le T$, and let $N_0\left(T\right)$ denote the number of non-trivial zeros of $\zeta \left(s\right)$ on the line $\alpha ={\displaystyle \frac{1}{2}},0<\beta \le T$. Selberg proved that there exist positive constants c and $T_0$, such that $N_0\left(T\right)>cN\left(T\right),(T>T_0)$ ^[11], later on, Levinson proved that $c\ge {\displaystyle \frac{1}{3}}$ ^[12], Lou and Yao proved that $c\ge 0.3484$ ^[13], Conrey proved that $c\ge {\displaystyle \frac{2}{5}}$ ^[14], Bui, Conrey and Young proved that $c\ge 0.41$ ^[15], Feng proved that $c\ge 0.4128$ ^[16], Wu proved that $c\ge 0.4172$ ^[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: ${\displaystyle \frac{1}{2}}\pm j14.1347251;{\displaystyle \frac{1}{2}}\pm j21.0220396;{\displaystyle \frac{1}{2}}\pm j25.0108575$.

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

$$1+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{3^2}}+{\displaystyle \frac{1}{4^2}}+{\displaystyle \frac{1}{5^2}}+\cdots ={\displaystyle \frac{{\pi }^2}{6}}$$

(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

$$\begin{array}{cc}\hfill {\displaystyle \frac{sinx}{x}}& =1-{\displaystyle \frac{x^2}{3!}}+{\displaystyle \frac{x^4}{5!}}-{\displaystyle \frac{x^6}{7!}}+\cdots =(1-{\displaystyle \frac{x^2}{{\pi }^2}})(1-{\displaystyle \frac{x^2}{4{\pi }^2}})(1-{\displaystyle \frac{x^2}{9{\pi }^2}})\cdots \hfill \end{array}$$

(9)

Then the author of this paper conjectured that $\xi \left(s\right)$ should be factored into $(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$ or something like that, which was verified by paring ${\rho }_i$ and ${\overline{\rho }}_i$ in the Hadamard product of $\xi \left(s\right)$, i.e., $(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})={\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$

The Hadamard product of $\xi \left(s\right)$ as shown in Equation (10) was first proposed by Riemann, however, it was Hadamard who showed the validity of this infinite product expansion ^[22].

$$\xi \left(s\right)=\xi \left(0\right)\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }})$$

(10)

where $\xi \left(0\right)={\displaystyle \frac{1}{2}}$, $\rho$ runs over all zeros of the completed zeta function $\xi \left(s\right)$.

Hadamard pointed out that to ensure the absolute convergence of the infinite product expansion, $\rho$ and $1-\rho$ are paired. Later in Section 3, we will show that $\rho$ and $\overline{\rho }$ 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 $\xi \left(s\right)$. And then we prove Lemma 3 to support the proof of the RH.

Multiple zeros of $\xi \left(s\right)$ and their real multiplicities: As shown in Figure 1, the multiple zeros of $\xi \left(s\right)$ are defined in terms of the quadruplet, i.e., $\rho ,\overline{\rho },1-\rho ,1-\overline{\rho }$.

There are two different expressions of factors of $\xi \left(s\right)/\xi (1-s)$ for the multiple zeros in Figure 1, respectively, i.e., $(1+{\displaystyle \frac{{(s-{\alpha }_1)}^2}{{\beta }_1^2}}{)}^2/(1+{\displaystyle \frac{{(1-s-{\alpha }_1)}^2}{{\beta }_1^2}}{)}^2$, or $(1+{\displaystyle \frac{{(s-{\alpha }_1)}^2}{{\beta }_1^2}})(1+{\displaystyle \frac{{(s-{\alpha }_2)}^2}{{\beta }_2^2}})/(1+{\displaystyle \frac{{(1-s-{\alpha }_1)}^2}{{\beta }_1^2}})(1+{\displaystyle \frac{{(1-s-{\alpha }_2)}^2}{{\beta }_2^2}})$ with ${\alpha }_1+{\alpha }_2=1,{\beta }_1^2={\beta }_2^2$.

To exclude the latter expression, we stipulate that zero ${\rho }_i$ related factors of $\xi \left(s\right)/\xi (1-s)$ take the unique form of $(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}/(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}$, where $m_i\ge 1$ is the real multiplicity of ${\rho }_i({\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$, here "real" means unique and unchangeable, with which we emphasize the objectivity of the multiplicity of ${\rho }_i({\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$. In Figure 1, the real multiplicity of ${\rho }_1({\overline{\rho }}_1,1-{\rho }_1,1-{\overline{\rho }}_1)$ is 2, i.e., $m_1=2$.

Remark: Although the real multiplicity $m_i$ of zero ${\rho }_i$ of $\xi \left(s\right)$ is unknown, it is an objective existence, unique, and unchangeable. This is the key point in the proof of Lemma 3.

Lemma 3: Given two absolutely convergent infinite products

$$f\left(s\right)=\prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}$$

(11)

and

$$f(1-s)=\prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}$$

(12)

where s is a complex variable, ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $\xi \left(s\right)$, $0<{\alpha }_i<1$ and ${\beta }_i\ne 0$ are real numbers, $m_i\ge 1$ is the real multiplicity of ${\rho }_i$, $0<|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|\le \cdots$.

Then we have

$$f\left(s\right)=f(1-s)\iff \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{1}{2}}\hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots \hfill \\ & i=1,2,3,\cdots ,\infty \hfill \end{array}\right.$$

(13)

where $"\iff "$ is the equivalent sign.

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

$$(1+{\displaystyle \frac{{(s-\alpha )}^2}{{\beta }^2}}{)}^m=(1+{\displaystyle \frac{{(1-s-\alpha )}^2}{{\beta }^2}}{)}^m\iff {(s-\alpha )}^2={(1-s-\alpha )}^2\iff \alpha ={\displaystyle \frac{1}{2}}$$

(14)

where $m\ge 1$ is positive integer, $\alpha \ne 0$ and $\beta \ne 0$ are real numbers.

Next, the proof is based on the divisibility of infinite products (or formal power series) with reference to the divisibility of polynomials. It is obvious that

$$\begin{array}{cc}\hfill f\left(s\right)=f(1-s)& \iff \prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=\prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & \iff (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l\left(s\right)=(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l(1-s)\hfill \end{array}$$

(15)

where

$$f_l\left(s\right)=\prod _{\begin{array}{c}i\in \mathbb{I}\setminus \{l\}\end{array}}(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}$$

(16)

$$f_l(1-s)=\prod _{\begin{array}{c}i\in \mathbb{I}\setminus \{l\}\end{array}}(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}$$

(17)

with $\mathbb{I}=\{1,2,3,\cdots ,\infty \}$, and "l" is an arbitrary element of set $\mathbb{I}$. In brief, $i\in \mathbb{I}\setminus \{l\}$ means that i runs over the elements of $\mathbb{I}$ excluding "l".

Then we have

$$\begin{array}{c}(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l\left(s\right)=(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l(1-s)\hfill \\ \Rightarrow \hfill \\ \left\{\begin{array}{cc}& (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l(1-s)\hfill \\ \hfill & (1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l\left(s\right)\hfill \end{array}\right.\hfill \end{array}$$

(18)

where "|" is the divisible sign.

Next, we exclude the possibility of $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|f_l(1-s)$ and $(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|f_l\left(s\right)$ in Equation (18) with the help of the real multiplicities of zeros of $\xi \left(s\right)$.

Considering $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}),0<{\alpha }_l<1,{\beta }_l\ne 0$, is irreducible over the field R of real numbers, we know from Equation (18) that

$$\begin{array}{cc}\hfill & (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|f_l(1-s)\Rightarrow (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}})|f_l(1-s)\hfill \\ \hfill & \Rightarrow \left(\mathrm{by}\mathrm{Lemma}6\right)\hfill \\ & (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}})|(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}),i\ne l\hfill \\ \hfill & \Rightarrow (1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}})=k(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}),i\ne l\hfill \\ \hfill & \Rightarrow \left(\mathrm{by}\mathrm{comparing}\mathrm{the}\mathrm{like}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{polynomial}\mathrm{equation}\right)\hfill \\ \hfill & {\alpha }_i+{\alpha }_l=1,{\beta }_i^2={\beta }_l^2,k=1,i\ne l\hfill \end{array}$$

Similarly,

$$\begin{array}{cc}\hfill & (1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|f_l\left(s\right)\Rightarrow (1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}})|f_l\left(s\right)\hfill \\ \hfill & \Rightarrow \left(\mathrm{by}\mathrm{Lemma}6\right)\hfill \\ & (1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}})|(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}),i\ne l\hfill \\ \hfill & \Rightarrow (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})=k(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}),i\ne l\hfill \\ \hfill & \Rightarrow \left(\mathrm{by}\mathrm{comparing}\mathrm{the}\mathrm{like}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{polynomial}\mathrm{equation}\right)\hfill \\ \hfill & {\alpha }_i+{\alpha }_l=1,{\beta }_i^2={\beta }_l^2,k=1,i\ne l\hfill \end{array}$$

As explained in the situation of Figure 1, ${\alpha }_i+{\alpha }_l=1,{\beta }_i^2={\beta }_l^2,i\ne l$ means that ${\alpha }_i\pm j{\beta }_i$ and ${\alpha }_l\pm j{\beta }_l$ are the same zeros in terms of quadruplet (${\rho }_i,{\overline{\rho }}_i,1-{\rho }_i$, and $1-{\overline{\rho }}_i$), which contradicts the definition of real multiplicities of zeros of $\xi \left(s\right)$.

Thus, in order to keep the real multiplicities of zeros of $\xi \left(s\right)$ unchanged, $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}$ can not divide $f_l(1-s)$, $(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}$ can not divides $f_l\left(s\right)$. In addition, $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}})$ is irreducible over the field R of real numbers, then by Lemma 8 we know that $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}$ and $f_l(1-s)$ are relative prime, similarly, $(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}$ and $f_l\left(s\right)$ are relative prime. Consequently, by Lemma 7, we obtain from Equation (18) the following result.

$$\begin{array}{cc}\hfill & (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l\left(s\right)=(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}{f}_l(1-s)\hfill \\ \hfill & \Rightarrow \hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}\hfill \\ \hfill & (1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}\hfill \\ \hfill & \Rightarrow \hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}=k(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}\hfill \\ \hfill & \Rightarrow (k=1,\mathrm{by}\mathrm{comparing}\mathrm{the}\mathrm{highest}-\mathrm{order}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{polynomial}\mathrm{equation})\hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}=(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}\hfill \\ \hfill & \Rightarrow \left(\mathrm{by}\mathrm{Equation}\left(14\right)\right)\hfill \\ \hfill & {\alpha }_l={\displaystyle \frac{1}{2}}\hfill \end{array}$$

(19)

Let l run over from 1 to ∞, and repeat the above process, we get

$$\begin{array}{cc}\hfill & \prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=\prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & \Rightarrow \hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & \Rightarrow \hfill \\ \hfill & {\alpha }_i={\displaystyle \frac{1}{2}},i=1,2,3,\cdots ,\infty \hfill \end{array}$$

(20)

Also, based on Equation (14), we have the following obvious fact

$$\begin{array}{cc}\hfill & {\alpha }_i={\displaystyle \frac{1}{2}},i=1,2,3,\cdots ,\infty \hfill \\ \hfill & \Rightarrow \hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & \Rightarrow \hfill \\ \hfill & \prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=\prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \end{array}$$

(21)

Further, limiting the imaginary parts ${\beta }_i$ of zeros to $0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots$ in order to keep the real multiplicities of zeros unchanged, we finally get

$$\begin{array}{cc}\hfill & \prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=\prod _{i=1}^{\infty }(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & \iff \hfill \\ \hfill & \left\{\begin{array}{cc}\hfill & (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots \hfill \\ \hfill & i=1,2,3,\cdots ,\infty \hfill \end{array}\right.\hfill \\ \hfill & \iff \hfill \\ \hfill & \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{1}{2}}\hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots \hfill \\ \hfill & i=1,2,3,\cdots ,\infty \hfill \end{array}\right.\hfill \end{array}$$

i.e.,

$$f\left(s\right)=f(1-s)\iff \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{1}{2}}\hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots \hfill \\ \hfill & i=1,2,3,\cdots ,\infty \hfill \end{array}\right.$$

That completes the proof of Lemma 3.

In order to support the proof of Lemma 3, we need the following classical results (Lemma 4 and Lemma 5) in Polynomial Algebra over Fields, with extension to infinite product (Lemma 6, Lemma 7, and Lemma 8).

To begin with, we introduce the ring of polynomial: $R\left[x\right]$, and the ring of formal power series: $R\left[\right[x\left]\right]$.

$R\left[x\right]$ is defined as the set of all polynomials in x over the field R of real numbers, i.e.,

$$R\left[x\right]=\{\sum _{i=0}^{\infty }{a}_i{x}^i|a_i\in R,a_i\ne 0\mathrm{for}\mathrm{all}\mathrm{but}\mathrm{a}\mathrm{finite}\mathrm{number}\mathrm{of}i\}$$

$R\left[\right[x\left]\right]$ is defined as the set of all formal power series (including infinite product of polynomial factors) in x over the field R of real numbers, i.e.,

$$R\left[\left[x\right]\right]=\{\sum _{i=0}^{\infty }{a}_i{x}^i|a_i\in R\}$$

The set $R\left[x\right]$ equipped with the operations + (addition) and · (multiplication) is the ring of polynomial in x over the field R. Similarly, $R\left[\right[x\left]\right]$ equipped with the operations + and · is the ring of formal power series in x over the field R. It is clear that $R\left[x\right]$ is a subset of $R\left[\right[x\left]\right]$, and that the algebraic operations of these two rings agree on this subset.

Lemma 4: Let $m\left(x\right),g_1\left(x\right),...,g_n\left(x\right)\in R\left[x\right],n\ge 2$. If $m\left(x\right)$ is irreducible (prime) and divides the product $g_1\left(x\right)\cdots g_n\left(x\right)$, then $m\left(x\right)$ divides one of the polynomials $g_1\left(x\right),...,g_n\left(x\right)$.

Lemma 5: Let $f\left(x\right),m\left(x\right)\in R\left[x\right]$. If $m\left(x\right)$ is irreducible and $f\left(x\right)$ is any polynomial, then either $m\left(x\right)$ divides $f\left(x\right)$ or $gcd(m,f)=1$, (gcd: greatest common divisor).

Lemma 6: Let $m\left(x\right),g_1\left(x\right),...,g_{\infty }\left(x\right)\in R\left[x\right]$. If $m\left(x\right)$ is irreducible and divides the product $g_1\left(x\right)\cdots g_{\infty }\left(x\right)\in R\left[\left[x\right]\right]$, then $m\left(x\right)$ divides one of the polynomials $g_1\left(x\right),...,g_{\infty }\left(x\right)$.

Lemma 7: Let $p_1\left(x\right),...,p_{\infty }\left(x\right),q\left(x\right),m\left(x\right)\in R\left[x\right],p\left(x\right)=p_1\left(x\right)\cdots p_{\infty }\left(x\right)\in R\left[\left[x\right]\right]$. If $m\left(x\right)$ is irreducible and divides the product $p\left(x\right)q\left(x\right)\in R\left[\right[x\left]\right]$, but $m\left(x\right)$ and $p\left(x\right)$ are relative prime, then $m\left(x\right)$ divides $q\left(x\right)$.

Lemma 8: Let $p_1\left(x\right),...,p_{\infty }\left(x\right),m\left(x\right)\in R\left[x\right],p\left(x\right)=p_1\left(x\right)\cdots p_{\infty }\left(x\right)\in R\left[\left[x\right]\right]$. If $m\left(x\right)$ is irreducible, then either $m\left(x\right)$ divides $p\left(x\right)$, or $m\left(x\right)$ and $p\left(x\right)$ are relative prime, i.e., $gcd(m,p)=1$.

Remark: The contents of Lemma 4 and Lemma 5 can be found in many textbooks of Linear Algebra, Modern Algebra, or Abstract Algebra, such as, Kenneth Hoffman, Ray Kunze, Linear Algebra (Second Edition), Prentice-Hall, INC., Englewood Cliffs, New Jersey, 1971; Linda Gilbert, Jimmie Gilbert, Elements of Modern Algebra (Seventh Edition), CENGAGE Learning, Belmont, CA, 2009; Henry C. Pinkham, Linear Algebra, Springer, 2015. Then we need only give the proofs of Lemma 6, Lemma 7, and Lemma 8.

Proof of Lemma 6: The proof is conducted by Transfinite Induction.

Let $P\left(\gamma \right)$ ($\gamma$ is an ordinal) be the statement of Lemma 4, i.e.,

"$m\left(x\right),g_1\left(x\right),...,g_n\left(x\right)\in R\left[x\right],n\ge 2$. If $m\left(x\right)$ is irreducible and divides the product $g_1\left(x\right)\cdots g_n\left(x\right)$, then $m\left(x\right)$ divides one of the polynomials $g_1\left(x\right),...,g_n\left(x\right)$" with n replaced by $\gamma$, where $\gamma \in A$, $A=\mathbb{N}\bigcup \{\omega \}$ with the ordering that $n<\omega$ for all natural numbers n, $\omega$ is the smallest limit ordinal other than 0.

Actually, Lemma 4 can be proved by Mathematical Induction, which includes the Base Case: $P\left(2\right)$ and the Successor Case:$P\left(n\right)\Rightarrow P(n+1)$ or $P\left(\gamma \right)\Rightarrow P(\gamma +1)$, of this proof.

Next we prove the Limit Case: $P(\gamma <\lambda )\Rightarrow P\left(\lambda \right)$, $\lambda$ is any limit ordinal other than 0.

For the sake of contradiction, assume that $P(\gamma <\lambda )\nRightarrow P\left(\lambda \right)$. Then, considering $m\left(x\right)$ is irreducible with the properties stated in Lemma 5, we have

$$\begin{gathered} m\left(x\right)|g_1\left(x\right)\cdots g_{\gamma }\left(x\right)\Rightarrow m\left(x\right)|g_1\left(x\right)\cdots g_{\gamma }\cdots g_{\lambda }\left(x\right)\Rightarrow gcd(m\left(x\right),g_i\left(x\right))=1,i\in \mathbb{N}\cup \{\omega ,\cdots ,\lambda \}\Rightarrow \\ gcd(m\left(x\right),g_i\left(x\right))=1,i\in \mathbb{N} \end{gathered}$$, which contradicts $P(\gamma <\lambda ):m\left(x\right)|g_1\left(x\right)\cdots g_{\gamma }\left(x\right)\Rightarrow m\left(x\right)\mathrm{divides}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{polynomials}{g}_1\left(x\right),...,g_{\gamma }\left(x\right)$.

Thus, we know that the assumption $P(\gamma <\lambda )\nRightarrow P\left(\lambda \right)$ does not hold.

Then $P(\gamma <\lambda )\Rightarrow P\left(\lambda \right)$ is true, i.e., the Limit Case is true.

That completes the proof of Lemma 6.

Proof of Lemma 7: If $m\left(x\right)$ is irreducible and divides the product $p\left(x\right)q\left(x\right)=p_1\left(x\right)\cdots p_{\infty }q\left(x\right)$, then, according to Lemma 6, $m\left(x\right)$ divides one of the polynomials $p_1\left(x\right),...,p_{\infty }\left(x\right),q\left(x\right)$. Further, if $m\left(x\right)$ and $p\left(x\right)$ are relative prime, then $m\left(x\right)$ does not divides any factor $p_i\left(x\right),i=1,\cdots ,\infty$ of $p\left(x\right)$ (otherwise $m\left(x\right)$ divides $p\left(x\right)$, which contradicts the condition "$m\left(x\right)$ and $p\left(x\right)$ are relative prime"). Thus, $m\left(x\right)$ must divides $q\left(x\right)$.

That completes the proof of Lemma 7.

Proof of Lemma 8:

Since $m\left(x\right)$ is irreducible, then by the definition of irreducible polynomial, either $gcd(m,p)=m\left(x\right)$ or $gcd(m,p)=1$. It is clear that $gcd(m,p)=m\left(x\right)\Rightarrow m\left(x\right)\mid p\left(x\right)$. Thus, we conclude that either $m\left(x\right)$ divides $p\left(x\right)$ or $gcd(m,p)=1$, i.e., $m\left(x\right)$ and $p\left(x\right)$ are relative prime.

That completes the proof of Lemma 8.

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 ${\alpha }_i={\displaystyle \frac{1}{2}},i=1,2,3,\cdots ,\infty$ in the new expression of $\xi \left(s\right)$ as shown in Equation (22).

Proof of the RH: The details are delivered in three steps as follows.

Step 1:

It is well-known that all the zeros of $\xi \left(s\right)$ always come in complex conjugate pairs. Then by pairing ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ in the Hadamard product as shown in Equation (10), we have

$$\begin{array}{cc}\hfill \xi \left(s\right)& =\xi \left(0\right)\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }})=\xi \left(0\right)\prod _{i=1}^{\infty }(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})\hfill \\ \hfill & =\xi \left(0\right)\prod _{i=1}^{\infty }(1-{\displaystyle \frac{s}{{\alpha }_i+j{\beta }_i}})(1-{\displaystyle \frac{s}{{\alpha }_i-j{\beta }_i}})=\xi \left(0\right)\prod _{i=1}^{\infty }({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})\hfill \end{array}$$

(22)

where $\xi \left(0\right)={\displaystyle \frac{1}{2}},0<{\alpha }_i<1,{\beta }_i\ne 0$.

The absolute convergence of the infinite product in Equation (22) in the form

$$\xi \left(s\right)=\xi \left(0\right)\prod _{i=1}^{\infty }(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})=\xi \left(0\right)\prod _{i=1}^{\infty }(1-{\displaystyle \frac{s(2{\alpha }_i-s)}{|{\rho }_i{|}^2}})$$

(23)

depends on the convergence of infinite series ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}$, which is an obvious fact according to Theorem 2 in Section 2, Chapter IV of Ref. ^[23].

Further, considering the absolute convergence of

$$\xi \left(s\right)=\xi \left(0\right)\prod _{i=1}^{\infty }(1-{\displaystyle \frac{s(2{\alpha }_i-s)}{|{\rho }_i{|}^2}})=\xi \left(0\right)\prod _{i=1}^{\infty }({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})$$

(24)

we have the following new expression of $\xi \left(s\right)$ by putting all the ${\rho }_i$ related multiple factors (zeros) together in the above Equation (24)

$$\xi \left(s\right)=\xi \left(0\right)\prod _{i=1}^{\infty }({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}}{)}^{m_i}$$

(25)

where $m_i\ge 1$ is the real multiplicity of ${\rho }_i$, $i=1,2,3,\cdots ,\infty$.

Step 2: Replacing s with $1-s$ in Equation (25), we obtain the infinite product expression of $\xi (1-s)$, i.e.,

$$\xi (1-s)=\xi \left(0\right)\prod _{i=1}^{\infty }{({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})}^{m_i}$$

(26)

Step 3: According to the functional equation $\xi \left(s\right)=\xi (1-s)$, and considering Equation (25) and Equation (26), we have

$$\xi \left(0\right)\prod _{i=1}^{\infty }{({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})}^{m_i}=\xi \left(0\right)\prod _{i=1}^{\infty }{({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})}^{m_i}$$

(27)

which is equivalent to

$$\prod _{i=1}^{\infty }{(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})}^{m_i}=\prod _{i=1}^{\infty }{(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}})}^{m_i}$$

(28)

where ${\beta }_i$ are in order of increasing $|{\beta }_i|$ , i.e., $0<|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|\le \cdots$.

To check the absolute convergence of both sides of Equation (28), it suffices to make a comparison with Equation (23) without considering multiple zeros in Equation (28), i.e., to make a comparison between ${\prod }_{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$ and $\xi \left(0\right){\prod }_{i=1}^{\infty }(1-{\displaystyle \frac{s(2{\alpha }_i-s)}{|{\rho }_i{|}^2}})$. It is well-known that the absolute convergence of $\xi \left(0\right){\prod }_{i=1}^{\infty }(1-{\displaystyle \frac{s(2{\alpha }_i-s)}{|{\rho }_i{|}^2}})$ depends on the convergence of infinite series ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}$ (already proved in Step 1); the absolute convergence of ${\prod }_{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$ depends on the convergence of infinite series ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}$, which is also an obvious fact because $0<{\alpha }_i<1,|{\rho }_i|\to \infty ,|{\beta }_i|\to \infty ,\mathrm{as}i\to \infty ,{lim}_{i\to \infty }{\displaystyle \frac{{\beta }_i^2}{|{\rho }_i{|}^2}}={lim}_{i\to \infty }{\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}=1$, that means ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}$ and ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}$ have the same convergence.

Then, according to Lemma 3, Equation (28) is equivalent to

$${\alpha }_i={\displaystyle \frac{1}{2}};0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots ;i=1,2,3,\cdots ,\infty$$

(29)

Thus, we conclude that all zeros of the completed zeta function $\xi \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$, 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 $\zeta \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

That completes the proof of the RH.

4. Conclusions

This paper presents a proof of the RH based on a new expression of $\xi \left(s\right)$, i.e.,

$$\xi \left(s\right)=\xi \left(0\right)\prod _{i=1}^{\infty }({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}}{)}^{m_i}$$

where $\xi \left(0\right)={\displaystyle \frac{1}{2}},{\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $\xi \left(s\right)$, $0<{\alpha }_i<1$ and ${\beta }_i\ne 0$ are real numbers, $0<|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|\le \cdots$, $m_i\ge 1$ is the real multiplicity of ${\rho }_i$.

The proof is conducted according to the divisibility implied in the (infinite) polynomial equation $\xi \left(s\right)=\xi (1-s)$. The first key-point is the paring of conjugate zeros $\rho$ and $\overline{\rho }$ to get the new expression of $\xi \left(s\right)$. The second key-point is the use of "real multiplicity" of a zero of $\xi \left(s\right)$. Obviously, the real multiplicity of a zero of $\xi \left(s\right)$ is an objective existence, unique, and unchangeable. As a result, the functional equation $\xi \left(s\right)=\xi (1-s)$ finally leads to ${\alpha }_i={\displaystyle \frac{1}{2}};0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots ;i=1,2,3,\cdots ,\infty$.