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

The completed zeta function (cid:24) ( s ) is expanded in MacLaurin series (inﬁnite polynomial), which can be further expressed as inﬁnite product (Hadamard product) of quadratic factors by its complex conjugate zeros (cid:11) i ± j(cid:12) i ; (cid:12) i ̸ = 0 ; i ∈ N ( N is the set of natural numbers, from 1 to inﬁnity). Then, according to the functional equation (cid:24) ( s ) = (cid:24) (1 − s ), we have (cid:24) ) which, by Lemma 3 and Corollary 1, is equivalent N with solution (cid:11) i = 12 ; i ∈ N (another solution s = 12 is invalid due to obvious contradiction). Thus, a proof of the Riemann Hypothesis is

which, by Lemma 3 and Corollary 1, is equivalent to with solution α i = 1 2 , i ∈ N (another solution s = 1 2 is invalid due to obvious contradiction). Thus, a proof of the Riemann Hypothesis is achieved.

Introduction and the problem description
It has been 162 years since the Riemann Hypothesis (RH) was proposed in 1859 [1] . Many efforts and achievements have been made towards proving this celebrated hypothesis, but it is still an open problem [2−3] .
The Riemann zeta function is the function of the complex variable s, defined in the half-plane ℜ(s) > 1 by the absolutely convergent series [2] ζ(s) = Riemann showed how to extend zeta function to the whole complex plane C by analytic continuation where θ(x) = ∑ ∞ −∞ e −n 2 πx 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 where γ is the Euler-Mascheroni constant. The connection between the Riemann zeta function and prime numbers can be established through the well-known Euler product.
the product being over the prime numbers p.
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 The Riemann zeta function ζ(s) has zeros at the negative even integers: −2, −4, −6, −8, · · · and one refers to them as the trivial zeros. The other zeros of ζ(s) are the complex numbers, i.e., non-trivial zeros [2] .
In 1896, Hadamard [4] and Poussin [5] independently proved that no zeros could lie on the line ℜ(s) = 1. Together with the functional equation and the fact that there are no zeros with real part greater than 1, this showed that all non-trivial zeros must lie in the interior of the critical strip 0 < ℜ(s) < 1. This was a key step in their first proofs of the famous Prime Number Theorem.
Later on, Hardy (1914) [6] , Hardy and Littlewood (1921) [7] showed that there are infinitely many zeros on the critical line ℜ(s) = 1 2 , which was an astonishing result at that time.
As further study, a completed zeta function ξ(s) is defined as 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 Eq.(6), and combining Eq.(5), we have the following functional equation Considering the definition of ξ(s), i.e., Eq. (10), and recalling Eq.(3), 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 ξ(s). Then we have the following Lemma 2.
According to Lemma 2, the following two statements for RH are equivalent.

Statement 1 of RH:
All the non-trivial zeros of ζ(s) have real part equal to 1 2 .

Statement 2 of RH:
All the zeros of ξ(s) have real part equal to 1 2 .
To prove the RH, the natural thinking is to estimate the number of zeros of ζ(s) in some certain closed areas according to the Argument Principle. Along this train of thought, there are some famous 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 ) > cN (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] .
Two types of infinite expansions of ξ(s), i.e., MacLaurin series (infinite polynomial) and infinite product (Hadamard product) by complex conjugate roots, will be adopted in this paper to open another door to the proof of RH.
The idea is originated from Euler's work on proving that This interesting and famous result is deduced mainly based on the abovementioned two types of infinite expansions, i.e., infinite polynomial and infinite product as follows This paper is also motivated by the fact that ξ(s) can be expressed by the following Hadamard product, which was first proposed by Riemann. However, it was Hadamard [17] who showed the validity of this infinite product expansion.
where ρ are precisely the non-trivial zeros of the Riemann zeta function ζ(s), or in another word, ρ runs over the zeros of the completed zeta function ξ(s). Besides, it is well-known that ξ(0) = 1 2 .

A proof of RH
This section is planned to give a proof of the Statement 2 and Statement 1 of RH. For this purpose, we need the following Lemma 3 and Corollary 1 on infinite product equations. Lemma 3: Given two infinite products and Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 9 December 2021 doi:10.20944/preprints202108.0146.v13 where s is a complex variable, α i , β i ̸ = 0, · · · ∈ R are all real numbers, i ∈ N are natural numbers, from 1 to infinity.
Then we have where " ⇔ " is the equivalent sign.
Proof: The proof is delivered by mathematical induction. First, it is obvious that Lemma 3 is true for i = 1, i.e., ( Second, suppose Lemma 3 is true for i = m, then we only need to prove Lemma 3 is true for i = m + 1. Thus we begin with the following Eq.(14) Now, let's consider i.e., Leaving out the special (trivial) cases which directly imply that Lemma 3 is true for i = m + 1, Eq. (16) is then equivalent to Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 9 December 2021 doi:10.20944/preprints202108.0146.v13 Without loss of generality, set where k is a real or complex number to be determined. Then Eq.(19) is equivalent to the following Eq.(20) Combining Eq.(21) and Eq. (14), we obtain i.e., Then we conclude that Lemma 3 is true for i = m + 1. Finally, by the principle of mathematical induction, Lemma 3 is true for all natural numbers i ∈ N, from 1 to infinity. That completes the proof of Lemma 3.

Corollary 1: Given two infinite products
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 9 December 2021 doi:10.20944/preprints202108.0146.v13 and where s is a complex variable, g(0) ̸ = 0, α i , β i ̸ = 0, · · · ∈ R are all real numbers, i ∈ N are natural numbers, from 1 to infinity. Then we have Proof: Based on the given conditions, it is obvious that Then, according to Lemma 3, we have That completes the proof of Corollary 1.
Next, we present the proof details of RH.

Proof of RH:
The details are delivered in three steps as follows.
Step 1: Since ξ(s) is an entire function, it is analytic in the whole complex plane C. Then ξ(s) can be expanded in MacLaurin series (infinite polynomial) at s = 0, i.e.
Thus, all the zeros of ξ(s) are the roots of the following infinite algebraic equation with real coefficients.
According to the well established theory of algebraic equation with real number coefficients, complex roots always come in pairs (complex conjugate pairs). Further by Lemma 2, all the zeros of ξ(s) are complex numbers, thus the roots of Eq.(29) can be denoted as complex conjugate pairs, i.e., Therefore, based on the Hadamard product Eq.(10), the infinite polynomial Eq.(28) can be rewritten as infinite product of quadratic factors by complex conjugate roots where ξ(0) = 1 2 .

Remark 1:
Actually, at first the author was inspired by Euler's work on proving Eq.(8) based on Eq.(9), and then speculated that ξ(s) could be expanded into quadratic factors Subsequently, the author starts with the Hadamard product Eq.(10) to get Eq.(30).
That completes the proof of the Statement 2 of RH.

Remark 2:
According to Lemma 2, we know that the Statement 1 of 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 .
Next, according to Lemma 3 and Corollary 1, the functional equation ξ(s) = ξ(1 − s) is equivalent to (s − α i ) 2 = (1 − s − α i ) 2 , i ∈ N with solution α i = 1 2 , i ∈ N. That means: 1) All the non-trivial zeros of the Riemann zeta function ζ(s) have real part equal to 1 2 ; 2) All the zeros of the completed zeta function ξ(s) have real part equal to 1 2 .
Then we conclude that the celebrated Riemann Hypothesis is true.