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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.

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1 Answer 1

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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.

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