# Riemann zeta function has no zeros on $\Re s=0,1$

This article shows that the Riemann zeta function  $\zeta(s)$ 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 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 functional equation

 $\pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}% \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)$

we have also that $\zeta(\sigma+it)\neq 0$ if

 $\sigma\leq\frac{c}{\log(\lvert t\rvert+1)}$

Bounding the zeros of $\zeta(s)$ away from $\Re s=0$, $1$ leads to a version of the prime number theorem with more precise error terms.

###### Theorem 1

$\zeta(1+it)\neq 0$ for $t\in\mathbb{R}$.

Proof. Notice that for $\theta\in\mathbb{C}$

 $0\leq 2(1+\cos\theta)^{2}=2\cos^{2}\theta+4\cos\theta+2=3+4\cos\theta+\cos(2\theta)$ (1)

If $\sigma=\Re s>1$, then $\zeta(\sigma+it)=\prod_{p\text{ prime}}(1-p^{-\sigma-it})^{-1}$, 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

 $\log\lvert\zeta(\sigma+it)\rvert=\sum_{p\text{ prime}}\sum_{m=1}^{\infty}\frac% {1}{mp^{m\sigma}}\cos(mt\log p)$

Using equation (1), we then have

 $\displaystyle 3\log\zeta(\sigma)+$ $\displaystyle 4\log\lvert\zeta(\sigma+it)\rvert+\log\lvert\zeta(\sigma+i2t)\rvert$ $\displaystyle=\sum_{p\text{ prime}}\sum_{m=1}^{\infty}\frac{1}{mp^{m\sigma}}(3% +4\cos(mt\log p)+\cos(2mt\log p))\geq 0$

so that

 $\zeta(\sigma)^{3}\lvert\zeta(\sigma+it)\rvert^{4}\lvert\zeta(\sigma+it\cdot 2)% \rvert\geq 1\ \text{ for all }\ \sigma>1,t\in\mathbb{R}$ (2)

But if $\zeta$ has a zero at $\sigma+it_{0}$, then

 $\lim_{\sigma\to 1^{+}}\zeta(\sigma)^{3}\lvert\zeta(\sigma+it_{0})\rvert^{4}% \lvert\zeta(\sigma+i2t)\rvert=0$

since the first factor gives a pole (http://planetmath.org/Pole) of order 3 at $1$ and the second factor gives a zero of order at least 4 at $1+it_{0}$. This contradicts equation (2).

###### Corollary 1

$\zeta(it)\neq 0$ for $t\in\mathbb{R}$.

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 $s=it$. The theorem implies that the RHS is nonzero, so the LHS is as well. Thus $\zeta(s)\neq 0$.

Title Riemann zeta function has no zeros on $\Re s=0,1$ RiemannZetaFunctionHasNoZerosOnReS01 2013-03-22 17:54:37 2013-03-22 17:54:37 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 11M06