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<r'<\frac{1}{4}r. Then

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



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