This article shows that $\zeta(s)$ has no zeros along the lines $\Re s=0$ or $\Re s=1$ . That implies that all nontrivial zeros of $\zeta(s)$ lie strictly within the critical strip $0<\Re s<1$ . As the parent article points out, this is known to be equivalent to one version of the prime number theorem.

It can in fact be shown that $\zeta(s)\neq 0$ for any $s=\sigma+it$ with $0<\sigma<1$ if $$ \sigma\geq 1-\frac{c}{\log(\lvert t\rvert+1) $$ for some constant $c$ . By using the root symmetry of $\zeta(s)$ , that also implies that $\zeta(\sigma+it)\neq 0$ if $$ \sigma \leq \frac{c}{\log(\lvert t\rvert+1) $$ This better set of estimates of locations of zeros of $\zeta(s)$ leads to a version of the prime number theorem with more precise error terms.

Proof. Notice that \begin{equation}\label{eqn:costheta} 0\leq 2(1+\cos\theta)^2 = 2\cos^2\theta + 4\cos \theta + 2 = 3+4\cos\theta+\cos(2\theta) \end{equation}If $\sigma=\Re s>1$ , then $\zeta(\sigma+it)=\prod_{p{ prime}} (1-p^{-z-it})^{-1}$ , so that $$ \log\zeta(\sigma+it)=-\sum_{p\text{ prime}} \log(1-p^{-\sigma-it}) = \sum_{p\text{ prime}} \sum_{m=1}^{\infty} \frac{1}{m}p^{-m\sigma-imt $$ and thus $$ \log\lvert\zeta(\sigma+it)\rvert = \sum_{p\text{ prime}}\sum_{m=1}^{\infty}\frac{1}{mp^{m\sigma}}\cos(mt\log p) $$ since the log of the absolute value is the real part of the log.

Using equation (), we then have

Proof. Use the functional equation $$ \pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s $$ and set $s=it$ .