# Asymptotic formula for the logarithmic derivative of zeta function

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.

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.