# Inequality for 1/zeta(s), zeta'(s)/zeta(s), and log zeta(s)

Theorem We have

$\displaystyle \zeta(s)=O(\log t)$

uniformly in the region

$\displaystyle 1-\frac{A}{\log t}\leq \sigma\leq 2,\ t>t_0$

where $A$ is any positive constant. In particular

$\displaystyle \zeta(1+it)=O(\log t).$

It is also easy to see that

$\displaystyle \zeta'(s)=O(\log^2 t)$

in the above region.

There is an alternative method, due to Landau obtaining results of this kind.

Lemma If $f(s)$ is analytic, and

$\displaystyle |f(s)|\leq |f(s_0)|e^M\quad (M>1)$

in the circle $|s-s_0|\leq r$, then

$\displaystyle |\frac{f'(s)}{f(s)}-\sum_{\rho}\frac{1}{s-\rho}|\leq \frac{AM}{r}\quad (|s-s_0|\leq \frac{1}{4}r),$

where $\rho$ runs through the zeros of $f(s)$ such that $|\rho-s_0|\leq \frac{1}{2}r$.

Lemma 1  If $f(s)$ satisfies the conditions of the previous lemma, and has no zeros in the right hand half of the circle $|s-s_0|\leq r$, then

$\displaystyle -\mathrm{Re}(\frac{f'(s_0)}{f(s_0)})\leq \frac{AM}{r};$

while if $f(s)$ has a zero $\rho_0$ between $s_0-\frac{1}{r}$ and $s_0$, then

$\displaystyle -\mathrm{Re}(\frac{f'(s_0)}{f(s_0)})\leq \frac{AM}{r}-\frac{1}{s_0-\rho_0}.$

Lemma Let $f(s)$ satisfy the conditions of Lemma 1, and let

$\displaystyle |\frac{f'(s_0}{f(s_0)}|\leq \frac{M}{r}.$

Suppose also that $f(s)\neq 0$ in the part $\sigma \geq\sigma_0-2r'$ of the circle $|s-s_0|\leq r$, where $0. Then

$\displaystyle |\frac{f'(s)}{f(s)}|\leq A\frac{M}{r}\quad (|s-s_0|\leq r').$