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

(more info)

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

(Theorem)

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

$\displaystyle \sigma\geq 1-\frac{c}{\log(\lvert t\rvert+1)}$

for some constant

. By using the root symmetry of $\zeta(s)$ , that also implies that $\zeta(\sigma+it)\neq 0$ if

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

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.

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

Proof. Notice that

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

(1)

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

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

and thus

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

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

Using equation (1), we then have

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