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 $\bar{\rho}$ $$\xi(s)= \xi(0)\prod_{i=1}^{\infty}(1-\frac{s}{\rho_i})(1-\frac{s}{\bar{\rho}_i})=\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}$$ 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, $\beta_i$ 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 $$\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(1-s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}$$ i.e., $$\prod_{i=1}^{\infty}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)}=\prod_{i=1}^{\infty}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}$$ 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.