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A Complete Proof Of The Riemann Hypothesis Based On A New Expression Of $\xi(s)$

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Submitted:

08 March 2022

Posted:

08 March 2022

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Abstract
Based on Hadamard product, a new expression of the completed zeta function $\xi(s)$ is obtained, 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 $\rho_i=\alpha_i+j\beta_i, \bar{\rho}_i=\alpha_i-j\beta_i$ are complex conjugate zeros of $\xi(s)$, $d_i\geq 1$ (natural numbers) are the multiplicities of $\rho_i$, $0<\alpha_i<1$ and $\beta_i\neq 0$ are real numbers, $i\in \mathbb{N}$ are natural numbers from 1 to infinity. $\beta_i$ are in order of increasing $|\beta_i|$, i.e., $|\beta_1|\leq|\beta_2|\leq|\beta_3|\leq, \cdots$. According to 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 $$\alpha_i= \frac{1}{2}, i\in \mathbb{N}, \text{ from 1 to infinity.}$$. Thus, we conclude that the Riemann Hypothesis is true.
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Subject: Computer Science and Mathematics  -   Algebra and Number Theory
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