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

Based on the Hadamard product \(\xi(s)= \xi(0)\prod_{\rho}(1-\frac{s}{\rho})\), a new expression of \(\xi(s)\) is obtained by paring \(\rho\) and \(\bar{\rho}\) \[\xi(s)=\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}^{d_{i}}\] where \(\xi(0)=\frac{1}{2}\), \(\rho_i=\alpha_i+j\beta_i\) and \(\bar{\rho}_i=\alpha_i-j\beta_i\) are 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 \(\rho_i\), \(\beta_i\) 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 equivalent to \[\alpha_i= \frac{1}{2}, i =1, 2, 3, \cdots, \infty\] Thus, we conclude that the Riemann Hypothesis is true.