The Riemann zeta function $\zeta(s)$ is defined by the formula
$$
\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}
$$
for $Re(s) > 1$.
The logarithmic derivative of the zeta function is
$$
-\frac{\zeta'}{\zeta}(s)=\sum_{n=1}^{\infty}\frac{\varLambda(n)}{n^s}
$$
for $Re(s) > 1$, where $\varLambda$ denotes the von Mangoldt function.
I've shown that the zeta function has a simple pole at $s=1$ with the resideu $1$.
Now,I am trying to show that $-\frac{\zeta'}{\zeta}(\sigma)\asymp \frac{1}{\sigma -1}$ for $1 < \sigma \leq 2$. That is,
$$
c\frac{1}{\sigma -1} \leq -\frac{\zeta'}{\zeta}(\sigma)\leq C\frac{1}{\sigma -1}
$$
for some contants $c,C$.
How can I show this inequality.
Best Answer
By partial summation we have for $\sigma>0$
$$ \zeta(s)=\frac{s}{s-1}-s\int_1^\infty \frac{f(u)}{u^{s+1}}\mathrm{d} u,$$
where $f(u)=u-\lfloor u\rfloor$, and hence $\zeta(s)=1+\frac{1}{s-1}+g(s)$ where $\lvert g(s)\rvert\leq \lvert s\rvert/\sigma$. Taking derivatives gives
$$\frac{\zeta'(s)}{\zeta(s)} = \frac{-(s-1)^{-2}+g'(s)}{1+(s-1)^{-1}+g(s)},$$
which is $\frac{-1}{s-1}+O(1)$ for $s=\sigma$ real and $3/4< \sigma <3$, say. This shows that
$$ -\frac{\zeta'}{\zeta}(\sigma) = \frac{1}{\sigma-1}+O(1)\quad\textrm{ for }\quad\frac{3}{4}<\sigma<3,$$
which is more than you require.