This is discussed before but I still have some gaps. I am referring to the answer Asymptotic formula for the logarithmic derivative of zeta function.
First we apply partial summation formula to $\zeta(s)= \sum_{n=1}^{\infty}1/n^s$.
The formula is (under suitable assumptions)
$\sum_{x<n\leq y} a_n \phi (n) = A(y) \phi(y) – A(x) \phi(x) – \int_{x}^{y} A(u) \phi'(u) du$ where
$A(t) = \sum_{0 \leq n \leq t} a_n $.
We should find $\zeta(s) = \frac{s}{s-1} – s\int_1^\infty (\sum_{0 \leq n \leq u} 1) \frac{1}{u^{s+1}} du$.
First I pick $\phi(n) = 1/n^s$ so $A(t) = \sum_{0 \leq n \leq t} 1$. In this way the integral part of formula is
$\int_{x}^{y} A(u) \phi'(u) du = -s\int_1^\infty (\sum_{0 \leq n \leq u} 1) \frac{1}{u^{s+1}} du$.
The sum $\sum_{0 \leq n \leq u} 1$ is the largest integer smaller than $u$, I guess we denote it as $u-\lfloor u\rfloor$. I could not find $\frac{s}{s-1}$.
At the end, we deduce $-\frac{\zeta'(s)}{\zeta(s)} = \frac{1}{\sigma -1} + O(1)$ for $\frac{3}{4} < \sigma < 3$.
But in my case I need $-\frac{\zeta'(c)}{\zeta(c)} \ll \log x $ where $c= 1+ (\log x)^{-1} $, $x \geq 2$. Can we conclude this result using $-\frac{\zeta'(s)}{\zeta(s)} = \frac{1}{\sigma -1} + O(1)$?
I appreciate any help!
("My case" is from Davenport's Multiplicative Number Theory page 107, I am working on the proof of explicit formula for $ \sum_{n \leq x} \Lambda(n)$ where it is necessary to understand the estimate I am asking).
Best Answer
Your linked answer is a bit sloppy. You need to prove that
$\zeta(s)-\frac1{s-1}$ is analytic for $\Re(s) >0$, using $\zeta(s)=\sum_{n\ge 1} \int_n^\infty s x^{-s-1}dx = \int_1^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1}+ s\int_1^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$
so that for $\Re(s)\in (0,1)$, $\zeta(s)= s\int_0^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$
whence $\zeta(s)$ doesn't vanish for $s\in (0,1)$. It is also obvious that it doesn't vanish for $s\in (1,\infty)$.
so that $f(s)=(s-1)\zeta(s)$ is analytic and it doesn't vanish for $s\in (0,\infty)$.
All together it means that $\frac{f'(s)}{f(s)}=\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1}$ is analytic on $(0,\infty)$, in particular $\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1}= O(1)$ on $[3/4,3]$