Residues of $\frac{x^s}{s}\frac{\zeta'(s)}{\zeta\phantom{'}(s)}$ Going through a proof in Analyti number theory, the calculation 
of the residues of
$$
    f(s) = \frac{x^s}{s}\frac{\zeta'(s)}{\zeta\phantom{'}(s)}
$$
came up. I do have some experience with complex analysis so I tried
to compute the residues myself. So far it seems the residues are
$$
\mathrm{Res}(0) = \log 2\pi \, , \
\mathrm{Res}(1) = -x \, , \
\mathrm{Res}(-2n) = -\frac{1}{n x^n}
$$
where $n \in \mathbb{N}$. I have been able to calculate the resiudes at $0$ and $1$ using the asymmetric functional equation for $\zeta(s)$. But alternative approaches is still very welcome.
My biggest problem is dealing with the residues at $-2n$, do anyone have any suggestions?
 A: Observe that poles and zeroes of $\zeta$ are all simple poles for its logarithmic derivative, hence the singularities of $s\longmapsto\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}$ lie all in halfplane $\{\sigma\leq1\}$.
Now, the poles of this function are: the pole of $\frac{x^{s}}{s}$ ($s=0$) which residue is
$$
Res\left(\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s},0\right)
=\lim_{s\rightarrow1}\left(s\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}\right)=\frac{\zeta'(0)}{\zeta(0)}=\log2\pi\;,
$$
and zeroes of $\zeta$ (trivial and non trivial) whose residue is 
\begin{align*}
Res\left(\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s},\rho\right)
=\lim_{s\rightarrow\rho}\left((s-\rho)\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}\right)=\frac{x^{\rho}}{\rho}\;,
\end{align*}
and the pole of $\zeta$ ($s=1$) which residue is
$$
Res\left(\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s},1\right)
=\lim_{s\rightarrow1}\left((s-1)\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}\right)=-x\;,
$$
in fact, since $(s-1)\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}=\frac{(s-1)^{2}\zeta'(s)}{(s-1)\zeta(s)}\frac{x^{s}}{s}$, being $s=1$ simple pole for $\zeta$, Laurent development around $1$ becomes $\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^{+\infty}c_{n}(s-1)^{n}$ which implies $\zeta'(s)=\frac{-1}{(s-1)^{2}}+\sum_{n=1}^{+\infty}nc_{n}(s-1)^{n-1}$. Then conclude by a simple computation.
