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

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14 April 2023

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17 April 2023

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

1. Introduction

It has been almost 164 years since the Riemann Hypothesis (RH) was proposed in 1859 [1]. Many efforts and achievements have been made towards proving this celebrated hypothesis, but it is still an open problem [2,3]. The Riemann zeta function is the function of the complex variable s, defined in the half-plane ( s ) > 1 by the absolutely convergent series [2]
ζ ( s ) = n = 1 1 n s , ( s ) > 1
The connection between the Riemann zeta function and prime numbers can be established through the well-known Euler product, i.e.
ζ ( s ) = n = 1 1 n s = p ( 1 p s ) 1 , ( s ) > 1
where p runs over the prime numbers.
Riemann showed how to extend zeta function to the whole complex plane C by analytic continuation
ζ ( s ) = π s / 2 Γ ( s / 2 ) { 1 s ( s 1 ) + 1 ( x s 2 1 + x s 2 1 2 ) · ( θ ( x ) 1 2 ) d x }
where θ ( x ) = e n 2 π x being the Jaccobi theta function, Γ being the Gamma function in the following Weierstrass expression (Meanwhile, there are also Gauss expression, Euler expression, and integral expression of the Gamma function.)
1 Γ ( s ) = s · e γ s n = 1 ( 1 + s n ) e s / n
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 s 2 Γ ( s 2 ) ζ ( s ) = π 1 s 2 Γ ( 1 s 2 ) ζ ( 1 s )
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 ( s ) = 1 . Together with the functional equation and the fact that there are no zeros with real part greater than 1, this showed that all non-trivial zeros must lie in the interior of the critical strip 0 < ( s ) < 1 . This was a key step in their first proofs of the famous Prime Number Theorem.
Later on, Hardy (1914) [6], Hardy and Littlewood (1921) [7] showed that there are infinitely many zeros on the critical line  ( s ) = 1 2 , which was an astonishing result at that time.
As a summary, we have the following results on the properties of the non-trivial zeros of ζ ( s ) [4,5,6,7,8,9].
Lemma 1. 
Non-trivial zeroes of ζ ( s ) , noted as ρ = α + j β , have the following properties
1) The number of non-trivial zeroes is infinity;
2) β 0 ;
3) 0 < α < 1 ;
4) ρ , ρ ¯ , 1 ρ ¯ , 1 ρ are all non-trivial zeroes.
As further study, a completed zeta function ξ ( s ) is defined as
ξ ( s ) = 1 2 s ( s 1 ) π s 2 Γ ( s 2 ) ζ ( s )
It is well-known that ξ ( s ) is an entire function of order 1. This implies ξ ( s ) is analytic, and can be expressed as infinite polynomial, in the whole complex plane C .
In addition, replacing s with 1 s in Equation (6), and combining Equation (5), we have the following functional equation
ξ ( s ) = ξ ( 1 s )
Considering the definition of ξ ( s ) , and recalling Equation (4), the trivial zeros of ζ ( s ) are canceled by the poles of Γ ( s 2 ) . The zero of s 1 and the pole of ζ ( s ) cancel; the zero s = 0 and the pole of Γ ( s 2 ) cancel [9,10]. Thus, all the zeros of ξ ( s ) are exactly the nontrivial zeros of ζ ( s ) . Then we have the following Lemma 2.
Lemma 2. 
The zeros of ξ ( s ) coincide with the non-trivial zeros of ζ ( s ) .
According to Lemma 2, the following two statements for the RH are equivalent.
Statement 1 of the RH: All the non-trivial zeros of ζ ( s ) have real part equal to 1 2 .
Statement 2 of the RH: All the zeros of ξ ( s ) have real part equal to 1 2 .
To prove the RH, a natural thinking is to estimate the numbers of non-trivial zeros of ζ ( s ) inside or outside some areas according to Argument Principle. Along this train of thought, there are many research works. Let N ( T ) denote the number of zeros of ζ ( s ) inside the rectangle: 0 < α < 1 , 0 < β T , and let N 0 ( T ) denote the number of zeros of ζ ( s ) on the line α = 1 2 , 0 < β 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 1 3 [12], Lou and Yao proved that c 0.3484 [13], Conrey proved that c 2 5 [14], Bui, Conrey and Young proved that c 0.41 [15], Feng proved that c 0.4128 [16], Wu proved that c 0.4172 [17].
On the other hand, many zeros have been calculated by hand or by computer programs. Among others, Riemann found the first three non-trivial zeros [18]. Gram found the first 15 zeros based on Euler-Maclaurin summation [19]. Titchmarsh calculated the 138th to 195th zeros using the Riemann-Siegel formula [20,21]. Here are the first three (pairs of) zeros: 1 2 ± j 14.1347251 ; 1 2 ± j 21.0220396 ; 1 2 ± j 25.0108575 .
The idea of this paper is originated from Euler’s work on proving the following famous equality
1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + = π 2 6
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
sin x x = 1 x 2 3 ! + x 4 5 ! x 6 7 ! + = ( 1 x 2 π 2 ) ( 1 x 2 4 π 2 ) ( 1 x 2 9 π 2 )
Then it is conjectured that ξ ( s ) should be factored into ( 1 + ( s α i ) 2 β i 2 ) or something like that, which was verified by paring ρ i and ρ ¯ i in the Hadamard product of ξ ( s ) to obtain
( 1 s ρ i ) ( 1 s ρ ¯ i ) = β i 2 α i 2 + β i 2 ( 1 + ( s α i ) 2 β i 2 )
The Hadamard product of ξ ( s ) as shown in Equation (10) was first proposed by Riemann, however, it was Hadamard [22] who showed the validity of this infinite product expansion.
ξ ( s ) = ξ ( 0 ) ρ ( 1 s ρ )
where ξ ( 0 ) = 1 2 , ρ runs over all the non-trivial zeros of the Riemann zeta function ζ ( s ) , or in another word, ρ runs over all the zeros of the completed zeta function ξ ( s ) .
To ensure the absolute convergence of the infinite product expansion, ρ and 1 ρ are paired. Later in Section 4, we will show that ρ and ρ ¯ can also be paired to ensure the absolute convergence of the infinite product expansion.

2. Lemmas

In this section, we first explain the concepts of multiple zeros of ξ ( s ) with their real multiplicities. And then we give three lemmas to support the proof of the RH, in which Lemma 3 is the key lemma.
Multiple zeros of ξ ( s ) : As shown in Figure 1, the multiple zeros of ξ ( s ) are defined in terms of the quadruplet, i.e., ρ , ρ ¯ , 1 ρ , 1 ρ ¯ .
It should be noticed that the multiple zeros with their real multiplicities of ξ ( s ) are objective existence, but the expression of the corresponding factors of ξ ( s ) are optional to some extent. For example, the multiple zeros as shown in Figure 1 have two different expressions as factors of ξ ( s ) and ξ ( 1 s ) , respectively, i.e., [ ( 1 + ( s α 1 ) 2 β 1 2 ) 2 , ( 1 + ( 1 s α 1 ) 2 β 1 2 ) 2 ] , or [ ( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) , ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 ) , α 1 + α 2 = 1 , β 1 2 = β 2 2 ] .
To exclude the latter expression, we stipulate that the multiple zeros ρ i related factors of ξ ( s ) take the form of ( 1 + ( s α i ) 2 β i 2 ) d i , where d i 1 is the real multiplicity of ρ i .
Lemma 3. 
Given two infinite products
f ( s ) = i = 1 ( 1 + ( s α i ) 2 β i 2 ) d i
and
f ( 1 s ) = i = 1 ( 1 + ( 1 s α i ) 2 β i 2 ) d i
where s is a complex variable, ρ i = α i + j β i and ρ ¯ i = α i j β i are the complex conjugate zeros of ξ ( s ) , 0 < α i < 1 and β i 0 are real numbers, d i 1 are the real multiplicities of ρ i , i are natural numbers from 1 to infinity, β i are in order of increasing | β i | , i.e., | β 1 | | β 2 | | β 3 | .
Then we have
f ( s ) = f ( 1 s ) α i = 1 2 , i = 1 , 2 , 3 , , | β 1 | < | β 2 | < | β 3 | <
where " " is the equivalent sign.
Proof. 
First of all, we have the following fact:
( 1 + ( s α ) 2 β 2 ) d = ( 1 + ( 1 s α ) 2 β 2 ) d ( s α ) 2 = ( 1 s α ) 2 α = 1 2
where d 1 is a natural number, α 0 and β 0 are real numbers.
Next, the proof is based on Transfinite Induction. Let P ( n ) be:
i = 1 n ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 n ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α n ) 2 β n 2 ) d n = ( 1 + ( 1 s α n ) 2 β n 2 ) d n | β 1 | < | β 2 | < | β 3 | < < | β n | α i = 1 2 , i = 1 n | β 1 | < | β 2 | < | β 3 | < < | β n |
According to Equation (14), P ( 1 ) is an obvious fact as the Base Case, i.e.,
i = 1 1 ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 1 ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 α 1 = 1 2
To be more convincing, let’s further check P ( 2 ) , i.e.,
i = 1 2 ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 2 ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 | β 1 | < | β 2 | α 1 = α 2 = 1 2 | β 1 | < | β 2 |
which is also an obvious fact according to Lemma 5.
As the Successor Case, we need to prove P ( n ) P ( n + 1 ) .
Actually, we have
i = 1 n + 1 ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 n + 1 ( 1 + ( 1 s α i ) 2 β i 2 ) d i i = 1 n ( 1 + ( s α i ) 2 β i 2 ) d i ( 1 + ( s α n + 1 ) 2 β n + 1 2 ) d n + 1 = i = 1 n ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( 1 s α n + 1 ) 2 β n + 1 2 ) d n + 1 ( by Lemma 5 ) i = 1 n ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 n ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( s α n + 1 ) 2 β n + 1 2 ) d n + 1 = ( 1 + ( 1 s α n + 1 ) 2 β n + 1 2 ) d n + 1 | β 1 | < | β 2 | < | β 3 | < < | β n | < | β n + 1 | ( by Equation ( 15 ) ) ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α n + 1 ) 2 β n + 1 2 ) d n + 1 = ( 1 + ( 1 s α n + 1 ) 2 β n + 1 2 ) d n + 1 | β 1 | < | β 2 | < | β 3 | < < | β n | < | β n + 1 | ( by Equation ( 14 ) ) α i = 1 2 , i = 1 , 2 , 3 , , n , n + 1 | β 1 | < | β 2 | < | β 3 | < < | β n | < | β n + 1 |
Thus the Successor Case is true, i.e., P ( n ) P ( n + 1 ) .
Next, we prove that P ( ) holds by considering well-ordered ordinal set A indexing the family of statements P ( γ : γ 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 ( ) holds, it suffices to prove the Limit Case, i.e., P ( γ < ω ) P ( ω ) .
Actually, we have
i = 1 ω ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 ω ( 1 + ( 1 s α i ) 2 β i 2 ) d i i = 1 γ < ω ( 1 + ( s α i ) 2 β i 2 ) d i ( 1 + ( s α ω ) 2 β ω 2 ) d ω = i = 1 γ < ω ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( 1 s α ω ) 2 β ω 2 ) d ω ( by Lemma 5 ) i = 1 γ < ω ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 γ < ω ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( s α ω ) 2 β ω 2 ) d ω = ( 1 + ( 1 s α ω ) 2 β ω 2 ) d ω | β 1 | < | β 2 | < | β 3 | < < | β ω | ( by P ( α < ω ) ) ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α ω ) 2 β ω 2 ) d ω = ( 1 + ( 1 s α ω ) 2 β ω 2 ) d ω | β 1 | < | β 2 | < | β 3 | < < | β ω | ( by Equation ( 14 ) ) α i = 1 2 , i = 1 , 2 , 3 , , ω | β 1 | < | β 2 | < | β 3 | < < | β ω |
Thus the Limit Case is true, i.e., P ( γ < ω ) P ( ω ) .
Hence we conclude by Transfinite Induction that P ( ) holds, i.e.,
i = 1 ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 ( 1 + ( 1 s α i ) 2 β i 2 ) d i ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α i ) 2 β i 2 ) d i = ( 1 + ( 1 s α i ) 2 β i 2 ) d i | β 1 | < | β 2 | < | β 3 | < α i = 1 2 , i = 1 | β 1 | < | β 2 | < | β 3 | <
i.e.,
f ( s ) = f ( 1 s ) α i = 1 2 , i = 1 , 2 , 3 , , | β 1 | < | β 2 | < | β 3 | <
That completes the proof of Lemma 3. □
Lemma 4. 
Given
( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 )
where s is a complex variable, 0 < α i < 1 and β i 0 are real numbers, | β 1 | | β 2 | .
Then we have
( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 ) α 1 = α 2 = 1 2 , | β 1 | | β 2 | α 1 + α 2 = 1 , | β 1 | = | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) , ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) , | β 1 | | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) , ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) , | β 1 | = | β 2 |
Proof. 
Expanding both sides of Equation (22), and comparing the coefficients of like terms, we obtain (details are omitted to save space)
( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 ) α 1 = α 2 = 1 2 , | β 1 | | β 2 | α 1 + α 2 = 1 , | β 1 | = | β 2 |
The inverse inference of Equation (24) is also an obvious fact. i.e.,
α 1 = α 2 = 1 2 , | β 1 | | β 2 | α 1 + α 2 = 1 , | β 1 | = | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 )
Then we have
( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 ) α 1 = α 2 = 1 2 , | β 1 | | β 2 | α 1 + α 2 = 1 , | β 1 | = | β 2 |
Further, according to Equation (14), i.e.,
( 1 + ( s α ) 2 β 2 ) d = ( 1 + ( 1 s α ) 2 β 2 ) d α = 1 2
and the following similar facts
( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) α 1 + α 2 = 1 , | β 1 | = | β 2 |
( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) α 1 + α 2 = 1 , | β 1 | = | β 2 |
we have
( 1 + ( s α 1 ) 2 β 1 2 ) ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) ( 1 + ( 1 s α 2 ) 2 β 2 2 ) α 1 = α 2 = 1 2 , | β 1 | | β 2 | α 1 + α 2 = 1 , | β 1 | = | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) , ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) , | β 1 | | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) , ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) , | β 1 | = | β 2 |
That completes the proof of Lemma 4. □
Remark 1. 
From the viewpoint of the products of (polynomial) factors, Lemma 4 reveals a fact that there are only two possible cases to make Eq. (22) hold, i.e., Case 1: ( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) , ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) ; and Case 2: ( 1 + ( s α 1 ) 2 β 1 2 ) = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) , ( 1 + ( s α 2 ) 2 β 2 2 ) = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) , respectively. This result is quite different from the similar equation of products of factors a · b = c · d , which has infinite cases of solutions for real numbers a , b , c , d .
Remark 2. 
Given
( 1 + ( s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2
where s is a complex variable, 0 < α 1 < 1 , 0 < α 2 < 1 and β 1 0 , β 2 0 are real numbers, d 1 1 , d 2 1 are natural numbers, denoting the real multiplicities of ρ 1 = α 1 + j β 1 and ρ 2 = α 2 + j β 2 , respectively, | β 1 | | β 2 | .
Then we have
( 1 + ( s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 α 1 = α 2 = 1 2 | β 1 | < | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 | β 1 | < | β 2 |
Proof. 
Based on Lemma 4, and considering additional possibilities that ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 , ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 (where "∣" is the divisible sign), or vice versa, we have
( 1 + ( s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 , ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 , | β 1 | < | β 2 | ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 , ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 d 1 = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 d 1 , | β 1 | = | β 2 | , d 1 < d 2 ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 , ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 d 2 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 d 2 , | β 1 | = | β 2 | , d 1 > d 2 ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 , ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 , | β 1 | = | β 2 | , d 1 = d 2 α 1 = α 2 = 1 2 , | β 1 | < | β 2 | α 1 = α 2 = 1 2 , | β 1 | = | β 2 | , d 1 < d 2 α 1 = α 2 = 1 2 , | β 1 | = | β 2 | , d 1 > d 2 α 1 + α 2 = 1 , | β 1 | = | β 2 | , d 1 = d 2 ( To exclude the duplication situations marked by between the first quadruplet of zeros represented by ρ 1 = α 1 + j β 1 and the sec ond quadruplet of zeros represented by ρ 2 = α 2 + j β 2 , which contradict the given condition that the real multiplicities of ρ 1 and ρ 2 are d 1 and d 2 , respectively . ) α 1 = α 2 = 1 2 | β 1 | < | β 2 | ( by Equation ( 14 ) ) ( 1 + ( s α 1 ) 2 β 1 2 ) d 1 = ( 1 + ( 1 s α 1 ) 2 β 1 2 ) d 1 ( 1 + ( s α 2 ) 2 β 2 2 ) d 2 = ( 1 + ( 1 s α 2 ) 2 β 2 2 ) d 2 | β 1 | < | β 2 |
That completes the proof of Lemma 5. □

3. A Proof of the RH

This section is planned to present a proof of the Riemann Hypothesis. We first prove that Statement 2 of the RH is true, and then by Lemma 2, Statement 1 of the RH is also true.
Proof of the RH. 
The details are delivered in three steps as follows.
Step 1: It is well-known that all the zeros of ξ ( s ) always come in complex conjugate pairs. Then by pairing ρ i = α i + j β i and ρ ¯ i = α i j β i in the Hadamard product as shown in Equation (10), we have
ξ ( s ) = ξ ( 0 ) ρ ( 1 s ρ ) = ξ ( 0 ) i = 1 ( 1 s ρ i ) ( 1 s ρ ¯ i ) = ξ ( 0 ) i = 1 ( 1 s α i + j β i ) ( 1 s α i j β i ) = ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( s α i ) 2 α i 2 + β i 2 )
where ξ ( 0 ) = 1 2 , 0 < α i < 1 , β i 0 .
The absolute convergence of the infinite product in Equation (30) in the form
ξ ( s ) = ξ ( 0 ) i = 1 ( 1 s ρ i ) ( 1 s ρ ¯ i ) = ξ ( 0 ) i = 1 ( 1 s ( 2 α i s ) | ρ i | 2 )
depends on the convergence of infinite series i = 1 1 | ρ i | 2 , or equivalently, ρ 1 | ρ | 2 , which is an obvious fact according to Theorem 2 in Section 2, Chapter IV of Ref. [23], as shown in the following.
Theorem 1. 
([23]).The function ξ ( s ) is an entire function of order one that has infinitely many zeros ρ n such that 0 Re ρ n 1 . The series | ρ n | 1 diverges, but the series | ρ n | 1 ε converges for any ε > 0 . The zeros of ξ ( s ) are the nontrivial zeros of ζ ( s ) .
Remark 3. 
In the Theorem 2 of Ref. [23], Re ( · ) is identical to ( · ) in this paper, both Re ( · ) and ( · ) mean the real part of any complex number.
Further, considering the absolute convergence of
ξ ( s ) = ξ ( 0 ) i = 1 ( 1 s ( 2 α i s ) | ρ i | 2 ) = ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( s α i ) 2 α i 2 + β i 2 )
we have the following new expression of ξ ( s ) by putting all the ρ i related multiple factors (zeros) together in the above Equation (32)
ξ ( s ) = ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( s α i ) 2 α i 2 + β i 2 ) d i
where d i 1 are the real multiplicities of ρ i , i are natural numbers from 1 to infinity.
Step 2: Replacing s with 1 s in Equation (33), we obtain the infinite product expression of ξ ( 1 s ) , i.e.,
ξ ( 1 s ) = ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( 1 s α i ) 2 α i 2 + β i 2 ) d i
Remark 4. 
According to the new expressions of ξ ( s ) and ξ ( 1 s ) , i.e., Equation (33) and Equation (34), we may conclude that all ρ i and 1 ρ i related multiple zeros, i.e., ( α i ± j β i ) d i , ( 1 α i ± j β i ) d i are included in the i t h group of factors, ( 1 + ( s α i ) 2 β i 2 ) d i and ( 1 + ( 1 s α i ) 2 β i 2 ) d i , respectively, or in another word, before or after the i t h group of factors of ξ ( s ) and ξ ( 1 s ) , there are no ρ i and 1 ρ i related multiple zeros.
Actually, with such arrangement of ρ i and 1 ρ i related multiple factors of ξ ( s ) and ξ ( 1 s ) , we "assigned" a reason for excluding, in the proof of Lemma 5 and Lemma 3, the "abnormal" situation, i.e., the successor factor and its predecessor factor represent the same quadruplet of zeros.
Step 3: According to the functional equation ξ ( s ) = ξ ( 1 s ) , and considering Equation (33) and Equation (34), we have
ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( s α i ) 2 α i 2 + β i 2 ) d i = ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( 1 s α i ) 2 α i 2 + β i 2 ) d i
which is equivalent to
i = 1 ( 1 + ( s α i ) 2 β i 2 ) d i = i = 1 ( 1 + ( 1 s α i ) 2 β i 2 ) d i
And that β i can be certainly arranged in order of increasing | β i | , i.e., | β 1 | | β 2 | | β 3 | .
Then, according to Lemma 3, Equation (36) is equivalent to
α i = 1 2 , i = 1 , 2 , 3 , , | β 1 | < | β 2 | < | β 3 | <
Thus, we conclude that all the zeros of the completed zeta function ξ ( s ) have real part equal to 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 1 2 .
That completes the proof of the RH. □
Remark 5. 
By Lemma 1, there are 2 pairs of complex zeros of ζ ( s ) simultaneously, i.e., ρ = α + j β , ρ ¯ = α j β , 1 ρ = 1 α j β , 1 ρ ¯ = 1 α + j β are all the non-trivial zeroes of ζ ( s ) . With the proof of the RH, i.e., α = 1 2 , these 2 pairs of zeros are actually only one pair, because ρ = 1 ρ ¯ = 1 2 + j β , ρ ¯ = 1 ρ = 1 2 j β . Thus Lemma 1 could be modified more precisely as follows.
Lemma 5. 
Non-trivial zeroes of ζ ( s ) , noted as ρ = α + j β , have the following properties
1) The number of non-trivial zeroes is infinity;
2) β 0 ;
3) 0 < α < 1 ;
4) ρ = 1 ρ ¯ , ρ ¯ = 1 ρ are all non-trivial zeroes.

4. Conclusion

The celebrated Riemann Hypothesis is proved to be true based on a new expression of the completed zeta function ξ ( s ) , i.e.,
ξ ( s ) = ξ ( 0 ) i = 1 ( β i 2 α i 2 + β i 2 + ( s α i ) 2 α i 2 + β i 2 ) d i
where ξ ( 0 ) = 1 2 , ρ i = α i + j β i and ρ ¯ i = α i j β i are the complex conjugate zeros of ξ ( s ) , 0 < α i < 1 and β i 0 are real numbers, d i 1 are the real multiplicities of ρ i , i are natural numbers from 1 to infinity, β i are in order of increasing | β i | , i.e., | β 1 | | β 2 | | β 3 | .

Acknowledgments

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

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Figure 1. Illustration of the multiple zeros of ξ ( s ) .
Figure 1. Illustration of the multiple zeros of ξ ( s ) .
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