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

1. Introduction

The RH $\left[1\right]$ is one of the most important unsolved problems in mathematics. Although there are many achievements towards proving this celebrated hypothesis, it remains an open problem [2,3]. The Riemann zeta function is originally defined in the half-plane $\Re (s)>1$ by the absolutely convergent series $\left[2\right]$

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

(1)

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

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

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

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

(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,

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

(3b)

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

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

(4)

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

As shown by Riemann, $\zeta (s)$ 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

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

(5)

The Riemann zeta function $\zeta (s)$ has zeros at the negative even integers: $-2$, $-4$, $-6$, $-8$, ⋯ and one refers to them as the trivial zeros. The other zeros of $\zeta (s)$ are the complex numbers, i.e., non-trivial zeros $\left[2\right]$.

In 1896, Hadamard $\left[4\right]$ and Poussin $\left[5\right]$ independently proved that no zeros could lie on the line $\Re (s)=1$, together with the functional equation $\xi (s)=\xi (1-s)$ 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 $0<\Re (s)<1$. Later on, Hardy (1914) $\left[6\right]$, Hardy and Littlewood (1921) $\left[7\right]$ showed that there are infinitely many zeros on the critical line $\Re (s)={\displaystyle \frac{1}{2}}$.

To give a summary of the related research works on the RH, we have the following results on the properties of the non-trivial zeros of $\zeta (s)$ [4–9].

Lemma 1: Non-trivial zeroes of $\zeta (s)$, 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 $\xi (s)$ is proposed by equation

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

(6)

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

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

(7)

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

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

Consequently, the following two statements are equivalent.

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

Statement 2: All zeros of $\xi (s)$ 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 $\zeta (s)$ 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 $\zeta (s)$ inside the rectangle: $0<\alpha <1,0<\beta \le T$, and let $N_0(T)$ denote the number of non-trivial zeros of $\zeta (s)$ 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(T)>cN(T),(T>T_0)$$\left[11\right]$, later on, Levinson proved that $c\ge {\displaystyle \frac{1}{3}}$$\left[12\right]$, Lou and Yao proved that $c\ge 0.3484$$\left[13\right]$, Conrey proved that $c\ge {\displaystyle \frac{2}{5}}$$\left[14\right]$, Bui, Conrey and Young proved that $c\ge 0.41$$\left[15\right]$, Feng proved that $c\ge 0.4128$$\left[16\right]$, Wu proved that $c\ge 0.4172$$\left[17\right]$.

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 $\left[18\right]$. Gram found the first 15 zeros based on Euler-Maclaurin summation $\left[19\right]$. 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 $\xi (s)$ 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 $\xi (s)$, 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 $\xi (s)$ as shown in Eq.(10) was first proposed by Riemann, however, it was Hadamard who showed the validity of this infinite product expansion $\left[22\right]$.

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

(10)

where $\xi (0)={\displaystyle \frac{1}{2}}$, $\rho$ runs over all zeros of $\xi (s)$.

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 multiplicity of a quadruplets of zeros of $\xi (s)/\xi (1-s)$. After that we prove Lemma 3 based on Lemmas 4-8. Lemma 3 is the key lemma to support the proof of the RH.

Multiple zeros of $\xi (s)/\xi (1-s)$: As shown in Figure 1, the multiple zeros of $\xi (s)$ are defined in terms of the quadruplets, i.e., $\rho ,\overline{\rho },1-\rho ,1-\overline{\rho }$.

If without any restriction, there are two different expressions of factors of $\xi (s)/\xi (1-s)$ for the multiple zeros in Figure 1, 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 ${\alpha }_1+{\alpha }_2=1,{\beta }_1^2={\beta }_2^2$.

The latter expression with ${\alpha }_1+{\alpha }_2=1,{\beta }_1^2={\beta }_2^2$ can be excluded with the use of multiplicity of zero, which is uniquely determined and then unchangeable, since $\xi (s)/\xi (1-s)$ is given. In Figure 1, the multiplicity of ${\rho }_1({\overline{\rho }}_1,1-{\rho }_1,1-{\overline{\rho }}_1)$ is 2, i.e., $m_1=2$.

Remark: Although the multiplicity $m_i$ of a quadruplets of zeros ${\rho }_i({\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$ of $\xi (s)/\xi (1-s)$ is unknown, it is an objective existence, finite, unique, and then unchangeable, for more details see Lemma 9 and Lemma 10. This is the key point in the proof of Lemma 3.

Lemma 3: Given two absolutely convergent infinite products

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

(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 $\xi (s)$, $0<{\alpha }_i<1$ and ${\beta }_i\ne 0$ are real numbers, $m_i\ge 1$ is the multiplicity of ${\rho }_i$, $0<|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|\le \cdots$.

Then we have

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

(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, $0<\alpha <1$ 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

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

(15)

where

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

(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

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

(18)

where "|" is the divisible sign.

We first exclude the possibility of $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|f_l(1-s)$ and $(1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}{)}^{m_l}|f_l(s)$ in Eq.(18) with the help of the uniqueness of the multiplicities of zeros of $\xi (s)$.

Considering the factor $(1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\beta }_l^2}}),0<{\alpha }_l<1,{\beta }_l\ne 0$, with discriminant $\Delta ={({\displaystyle \frac{2{\alpha }_l}{{\beta }_l^2}})}^2-4·{\displaystyle \frac{1}{{\beta }_l^2}}(1+{\displaystyle \frac{{\alpha }_l^2}{{\beta }_l^2}})=-4·{\displaystyle \frac{1}{{\beta }_l^2}}<0$, is irreducible over the field R of real numbers, we know from Eq.(18) that

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

Similarly, we also have

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

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 uniqueness of the multiplicities of zeros of $\xi (s)$.

Thus, in order to keep the multiplicities of zeros of $\xi (s)$ 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 $f_l(s)$. 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 $f_l(s)$ are relative prime. Consequently, by Lemma 7, we obtain from Eq.(18) the following result.

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

(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 Eq.(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 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.,

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

That completes the proof of Lemma 3.

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 of polynomial factors as a special expression of formal power series (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 as a special case) 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 $m(x),g_1(x),...,g_n(x)\in R\left[x\right],n\ge 2$. If $m(x)$ is irreducible (prime) and divides the product $g_1(x)\cdots g_n(x)$, then $m(x)$ divides one of the polynomials $g_1(x),...,g_n(x)$.

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

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

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

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

Remark: The contents of Lemma 4 and Lemma 5 can be found in many textbooks of linear algebra, modern algebra, or abstract algebra, see for example references [24–26]. As to the contents related to formal power series, see references [27,28]. Below we give the proofs of Lemma 6, Lemma 7, and Lemma 8.

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

Let $P(\gamma )$ ($\gamma$ is an ordinal number) be the statement:

"$m(x),g_1(x),...,g_{\gamma }(x)\in R\left[x\right],\gamma \ge 2$. If $m(x)$ is irreducible and divides the product $g_1(x)\cdots g_{\gamma }(x)$, then $m(x)$ divides one of the polynomials $g_1(x),...,g_{\gamma }(x)$", 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.

Base Case: $P(2)$ is an obvious fact according to Lemma 4 with $n=2$;

Successor Case: To prove $P(\gamma )\Rightarrow P(\gamma +1)$, we have $g_1(x)\cdots g_{\gamma }(x)g_{\gamma +1}(x)\iff g(x)·g_{\gamma +1}(x)$, where $g(x)=g_1(x)\cdots g_{\gamma }(x)$. Then according to Lemma 4 with $n=2$, we have $m(x)\mid g(x)·g_{\gamma +1}(x)\Rightarrow m(x)\mid g(x)$ or $m(x)\mid g_{\gamma +1}(x)$. Considering $P(\gamma )$: if $m(x)$ divides $g(x)$, then $m(x)$ divides one of $g_1(x),\cdots ,g_{\gamma }(x)$, thus we know $P(\gamma )\Rightarrow P(\gamma +1)$.

Limit Case: We need to prove $P(\gamma <\lambda )\Rightarrow P(\lambda )$, $\lambda$ is any limit ordinal other than 0. For the sake of contradiction, assume that $P(\gamma <\lambda )\nRightarrow P(\lambda )$. Then, considering $m(x)$ is irreducible with the properties stated in Lemma 5, we have: $m(x)|g_1(x)\cdots g_{\gamma }(x)\Rightarrow m(x)|g_1(x)\cdots g_{\gamma }\cdots g_{\lambda }(x)\Rightarrow gcd(m(x),g_i(x))=1,1\le i\le \lambda \Rightarrow gcd(m(x),g_i(x))=1,i\in \mathbb{N}$, which contradicts $P(\gamma <\lambda ):m(x)|g_1(x)\cdots g_{\gamma }(x)\Rightarrow m(x)\mathrm{divides}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{polynomials}{g}_1(x),...,g_{\gamma }(x),\gamma \in \mathbb{N}$.

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

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

That completes the proof of Lemma 6.

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

That completes the proof of Lemma 7.

Proof of Lemma 8:

Since $m(x)$ is irreducible, then by the definition of irreducible polynomial, either $gcd(m(x),p(x))=k·m(x),k\in \mathbb{R},k\ne 0$ or $gcd(m(x),p(x))=1$. It is clear that $gcd(m(x),p(x))=k·m(x)\Rightarrow m(x)\mid p(x)$. Thus, we conclude that either $m(x)$ divides $p(x)$ or $gcd(m(x),p(x))=1$, i.e., $m(x)$ and $p(x)$ are relative prime.

That completes the proof of Lemma 8.

Additionally, we also need the following results on properties of a zero of entire function in complex analysis for understanding the multiplicity of a zero of $\xi (s)$.

Lemma 9: The multiplicity of a zero of any non-zero entire function is a finite positive integer.

Proof: Let $f(s)\not\equiv 0,s\in \mathbb{C}$, be an entire function, which means it is holomorphic on the whole complex plane. Suppose $f(s)$ has a zero at $s_0\in \mathbb{C}$ of multiplicity m, then $f(s)={(s-s_0)}^{m}g(s)$, where $g(s)$ is also an entire function and $g(s_0)\ne 0$.

Assume for contradiction that m is infinite, which implies there exists an accumulation point of zeros in the neighbor of $s_0$. Then, by Identity Theorem for holomorphic functions, and considering "0" is also an entire function, we have $f(s)\equiv 0,s\in \mathbb{C}$, which contradicts the given condition that $f(s)\not\equiv 0,s\in \mathbb{C}$. Thus, the assumption is false, i.e., m must be a finite positive integer.

That completes the proof of Lemma 9.

Lemma 10: The multiplicity of a zero of any non-zero entire function is unique.

Proof: Let $f(s)\not\equiv 0,s\in \mathbb{C}$, be an entire function, which has a multiple zero at $s_0\in \mathbb{C}$ of multiplicity m. We can write: $f(s)={(s-s_0)}^{m}g(s)$, where $g(s)$ is also an entire function and $g(s_0)\ne 0$.

Assume for contradiction that there exists another integer $n\ne m$ such that n is also a multiplicity of the zero $s_0$. This means we can also write: $f(s)={(s-s_0)}^{n}h(s)$, where $h(s)$ is an entire function and $h(s_0)\ne 0$.

Since both expressions for $f(s)$ must be equal, we then obtain ${(s-s_0)}^{m}g(s)={(s-s_0)}^{n}h(s)$. Without loss of generality, consider $m>n$, then we have: ${(s-s_0)}^{m-n}g(s)=h(s)\Rightarrow h(s_0)=0$, which is a contradiction to $h(s_0)\ne 0$. Thus, the assumption is false, i.e., the multiplicity of a zero of any non-zero entire function is unique.

That completes the proof of Lemma 10.

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 $\xi (s)$ as shown in Eq.(22).

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

Step 1:

It is well-known that zeros of $\xi (s)$ 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 Eq.(10), we have

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

(22)

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

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

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

(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

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

(24)

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

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

(25)

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

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

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

(26)

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

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

(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 Eq.(28), it suffices to make a comparison with Eq.(23) without considering multiple zeros in Eq. (28), i.e., to make a comparison between ${\prod }_{i=1}^{\infty }(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$ and $\xi (0){\prod }_{i=1}^{\infty }(1-{\displaystyle \frac{s(2{\alpha }_i-s)}{|{\rho }_i{|}^2}})$. It is well-known that the absolute convergence of $\xi (0){\prod }_{i=1}^{\infty }(1-{\displaystyle \frac{s(2{\alpha }_i-s)}{|{\rho }_i{|}^2}})$ 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, Eq.(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 $\xi (s)$ 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 $\zeta (s)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

That completes the proof of the RH.

4. Conclusion

This paper presents a proof of the RH based on a new expression of $\xi (s)$, i.e., $\xi (s)=\xi (0){\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 $\xi (0)={\displaystyle \frac{1}{2}},{\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $\xi (s)$, $0<{\alpha }_i<1$ 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 multiplicity of ${\rho }_i$.

The proof is conducted with the help of the divisibility contained in the functional equation $\xi (s)=\xi (1-s)$ expressed as infinite products of polynomial factors. The first key-point is the paring of conjugate zeros $\rho$ and $\overline{\rho }$ to get the new expression of $\xi (s)$. The second key-point is the use of multiplicity of zero. Obviously, the multiplicity of a zero of $\xi (s)$ is an objective existence, uniquely determined, and then unchangeable, although unknown. As a result, the functional equation $\xi (s)=\xi (1-s)$ finally leads to ${\alpha }_i={\displaystyle \frac{1}{2}};0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\cdots ;i=1,2,3,\cdots ,\infty$.