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

The completed zeta function \(\xi(s)\) 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 \(\rho_i=\alpha_i +j\beta_i, \bar{\rho}_i=\alpha_i-j\beta_i, 0<\alpha_i<1, \beta_i\neq 0, i\in \mathbb{N}\) are natural numbers from 1 to infinity, \(\rho_i\) are in order of increasing $|\rho_i|=\sqrt{\alpha_i^2+\beta_i^2}$, i.e., $|\rho_1|<|\rho_2|\leq|\rho_3|\leq |\rho_4|, \cdots$, together with \(\beta_1<\beta_2\leq\beta_3\leq\beta_4, \cdots\). Then, according to the functional equation \(\xi(s)=\xi(1-s)\), we have \[\xi(0)\prod_{i\in \mathbb{N}}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)} =\xi(0)\prod_{i\in \mathbb{N}}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(1-s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}\] which, by Lemma 3, is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}, \text{from 1 to infinity.}$$ with only valid solution \(\alpha_i= \frac{1}{2}\) (another solution \(s=\frac{1}{2}\) is invalid due to obvious contradiction). Thus, a proof of the Riemann Hypothesis is achieved.