Riemann zeta function has no zeros on $\mathrm{\Re }s=0,1$

This article shows that the Riemann zeta function $\zeta (s)$ has no zeros along the lines $\mathrm{\Re }s=0$ or $\mathrm{\Re }s=1$. That implies that all nontrivial zeros of $\zeta (s)$ lie strictly within the critical strip . 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 $\zeta (s)\ne 0$ for any $s=\sigma +it$ with if

$$\sigma \ge 1-\frac{c}{\log (|t|+1)}$$

for some constant $c$. By using the functional equation

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

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

$$\sigma \le \frac{c}{\log (|t|+1)}$$

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

Theorem 1

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

Proof. Notice that for $\theta \in ℂ$

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

If $\sigma =\mathrm{\Re }s>1$, then $\zeta (\sigma +it)={\prod }_{p\text{ prime}}{(1-p^{-\sigma -it})}^{-1}$, so that

$$\log \zeta (\sigma +it)=-\sum _{p\text{ prime}}\log (1-p^{-\sigma -it})=\sum _{p\text{ prime}}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m}{p}^{-m\sigma -imt}$$

and thus

$$\log |\zeta (\sigma +it)|=\sum _{p\text{ prime}}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m{p}^{m\sigma }}\cos (mt\log p)$$

since the log of the absolute value is the real part of the log.

Using equation (1), we then have

$3\log \zeta (\sigma )+$$4\log |\zeta (\sigma +it)|+\log |\zeta (\sigma +i2t)|$

$={\displaystyle \sum _{p\text{ prime}}}{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m{p}^{m\sigma }}}(3+4\cos (mt\log p)+\cos (2mt\log p))\ge 0$

so that

$$\zeta {(\sigma )}^3{|\zeta (\sigma +it)|}^4|\zeta (\sigma +it\cdot 2)|\ge 1\text{ for all }\sigma >1,t\in ℝ$$(2)

But if $\zeta$ has a zero at $\sigma +i{t}_0$, then

$$\underset{\sigma \to 1^{+}}{lim}\zeta {(\sigma )}^3{|\zeta (\sigma +i{t}_0)|}^4|\zeta (\sigma +i2t)|=0$$

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 $1+i{t}_0$. This contradicts equation (2).

Corollary 1

$\zeta (it)\ne 0$ for $t\mathrm{\in }\mathrm{R}$.

Proof.  Use the functional equation

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

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