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

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Submitted:

13 August 2022

Posted:

15 August 2022

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Abstract
A new expression of the completed zeta function $\xi(s)$ is obtained according to the Hadamard product, 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 $\xi(s)$, $0<\alpha_i<1$ and $\beta_i\neq 0$ are real numbers, $d_i\geq 1$ are the multiplicities of $\rho_i$, $i$ are natural numbers from 1 to infinity, $\beta_i$ are in order of increasing $|\beta_i|$, i.e., $|\beta_1|\leq|\beta_2|\leq|\beta_3|\leq \cdots$. Then we have, by the functional equation $\xi(s)=\xi(1-s)$, that $$\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}} =\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(1-s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}}$$ 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 further equivalent to $$(s-\alpha_i)^2=(1-s-\alpha_i)^2 \Leftrightarrow \alpha_i= \frac{1}{2}, \text{with } i\text{ from 1 to infinity.}$$ 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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