Find the residue of $\frac{\zeta'(1+s)}{\zeta(1+s)}\frac{x^s}{s}$ at $s=0$

Question

Let, $\displaystyle f(s)=\frac{\zeta'(1+s)}{\zeta(1+s)}\frac{x^s}{s}$. Prove that $Res(f,s=0)=A-\log x$ , for some constant $A$.

At $s=0$ , $\zeta(s)$ has a pole of order $1$ and $\zeta'(s)$ has a pole of order $2$. So, $\frac{\zeta'(1+s)}{\zeta(1+s)}$ has a simple pole at $s=0$. Again , $x^s/s$ has a simple pole at $s=0$. So $f$ has a pole of order $2$ at $s=0$. So the residue is given by

$$\lim_{s\to 0}\frac{d}{ds}\left\{sx^s.\frac{\zeta'(1+s)}{\zeta(1+s)}\right\}.$$

I've stuck here ! How can I calculate this to prove the result ?

Answer 1

You could get an easy access to answer if you already know the Stieltjes constants which appears in the Laurent series expansion of $\zeta(s)$

$$\displaystyle \zeta (s)={\frac{1}{s-1}}+\sum _{n=0}^{\infty}{\frac{(-1)^{n}}{n!}}\gamma _{n}(s-1)^{n}$$

and you take the first and constant item

$$\displaystyle \zeta (s+1)={\frac{1}{s}}+\gamma_{0}+O(s)$$

then you can find

$$\displaystyle \begin{align} \frac{\zeta'(1+s)}{\zeta(1+s)}&=\frac{d\ln(\zeta(1+s))}{ds}\\&=\frac{d}{ds}\left(\ln(1+\gamma_{0}s+O(s^2))-\ln s\right)=\frac{d}{ds}\left(\gamma_{0}s+O(s^2)-\ln s\right)\\&={-\frac{1}{s}}+\gamma_{0}+O(s) \end{align}$$

notice you have

$$\displaystyle \frac{x^s}{s}=\frac{1}{s}+\ln x+O(s)$$

so you can easily get

$$\displaystyle f(s)=-\frac{1}{s^2}+\frac{\gamma_{0}-\ln x}{s}+O(1)$$

here $\gamma_{0}$ is Euler–Mascheroni constant, and the simply proof on the constant appears in the series can see another question like this one.

Answer 2

Due to the integral representation for the $\zeta$ function, in a neighbourhood of $s=0$ we have $$ \zeta(s+1)=\frac{1}{s}+\gamma+(\gamma_1+o(1))s = \frac{1}{s}\left[1+\gamma s+O(1)s\right]\tag{1} $$ hence by applying $\frac{d}{ds}\log(\cdot)$ to both sides $$ \frac{\zeta'(s+1)}{\zeta(s+1)} = -\frac{1}{s}+\gamma+O(1)s \tag{2} $$ while $$ \frac{x^s}{s} = \frac{1}{s}\exp(s\log x)=\frac{1}{s}+\log x+O(1)s\log^2(x)\tag{3}$$ hence by multiplying $(2)$ and $(3)$ we get that $s=0$ is a double pole of $\frac{\zeta'(s+1)}{\zeta(s+1)}\cdot\frac{x^s}{s}$ with residue $\gamma-\log x$.