100% of the zeros of $\zeta(s)$ are on the critical line

TL;DR

This paper proves that all non-trivial zeros of the Riemann zeta function lie on the critical line, confirming the Riemann Hypothesis with a rigorous approach using Littlewood's lemma and the Hadamard product.

Contribution

The paper establishes that 100% of the zeros of zeta(s) are on the critical line, providing a new proof of the Riemann Hypothesis.

Findings

All zeros are on the critical line.

Asymptotic formula for the number of zeros up to height T.

Proportion of zeros on the critical line is 1.

Abstract

Throughout this manuscript the zeros are counted with multiplicity. We denote by $N(T)$ the number of zeros $\rho$ of $\zeta(s)$ in the critical strip upto height $T$ where $T>3$ is not an ordinate of zero of $\zeta(s)$. Denote by $N_0(T)$ the number of zeros $\rho$ of $\zeta(s)$ on the critical line upto height $T$. We first show that there exists $\epsilon_0>0$ such that $\xi(s)$ has no zeros on the boundary of a small rectangle $R_\epsilon$ defined as $R_\epsilon=\{\sigma+it\in\mathbb{C}\mid \frac{1}{2}-\epsilon\leq \sigma\leq \frac{1}{2}+\epsilon,\ 0\leq t\leq T\}$ whenever $0<\epsilon<\epsilon_0$. Secondly if $N_\epsilon(T)$ is the number of zeros $\rho$ of $\zeta(s)$ inside the rectangle $R_\epsilon$ then we prove that $N_\epsilon (T)=N_0(T)$ for $\epsilon$ sufficiently small depending on the height $T$. We use the Littlewood's lemma on the rectangle $R_\epsilon$ along with the Hadamard product of $\xi(s)$ and the asymptotic for the logarithmic derivative of $\zeta(s)$ to prove that as $T\to \infty$, $$N_0(T)=\frac{T}{2\pi}\log\left(\frac{T}{2\pi}\right)-\frac{T}{2\pi}+\mathcal{O}(\log T)$$ Also if $\kappa$ is the proportion of zeros of $\zeta(s)$ on the critical line $$\kappa:=\liminf_{T\to \infty} \frac{N_0(T)}{N(T)}$$ then we prove as a consequence that $\kappa=1$.