Calculating the residues of $\frac{\zeta^{\prime}{(s) x^{s}}}{\zeta(s)\cdot s}$ Calculating the poles of $\frac{\zeta^{\prime}{(s) x^{s}}}{\zeta(s)\cdot s}$, where x is a fixed real number. 
I am trying to calculate the poles of this function at the trivial zeros of $\zeta$, namely the even negative integers. 
To do so I image one looks at 
$\lim_{s \to \ -2k} \frac{(s+2k)\zeta^{\prime}{(s) x^{s}}}{\zeta(s)\cdot s}$
but I have trouble when evaluating the $\zeta^{\prime}$ at $-2k$.
Is my strategy bad or am I able to calculate $\zeta^{\prime}(-2k)$ for every k? I suspect the value to be 1. 
This is almost exactly the same question, but no satisfying answer has been given. 
 A: For a fixed $x\in \mathbb{R}\setminus \{0\}$, $s\mapsto x^s$ is an entire function (you can choose different branches of that function by choosing different logarithms of $x$, but they all are entire), so that factor contributes no poles. In simple poles of the function, it modifies the residue just by multiplication with the value of $x^s$ in the pole.
The factor $\frac{1}{s}$ contributes a simple pole in $0$ (since $\zeta(0)\neq 0$), and the residue
$$\operatorname{Res}\left(\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s}; 0\right) = \frac{\zeta'(0)}{\zeta(0)} = \frac{-\frac12\log (2\pi)}{-\frac12} = \log (2\pi)$$
is easy to find.
So the poles arising from $\frac{\zeta'(s)}{\zeta(s)}$ remain. Generally, if $f$ is holomorphic in a punctured neighbourhood of $z_0$ and has a pole or zero in $z_0$, we have a representation $f(z) = (z-z_0)^k\cdot h(z)$ in that punctured neighbourhood, where $h$ is holomorphic with $h(z_0) \neq 0$. Then the logarithmic derivative of $f$ in that punctured neighbourhood is
$$\frac{f'(z)}{f(z)} = \frac{k(z-z_0)^{k-1}\cdot h(z) + (z-z_0)^k\cdot h'(z)}{(z-z_0)^k\cdot h(z)} = \frac{k}{z-z_0} + \frac{h'(z)}{h(z)}.$$
Since $\frac{h'}{h}$ is holomorphic in a neighbourhood of $z_0$, we see that $\frac{f'}{f}$ has a simple pole in $z_0$ with residue $k$ [so the residue is the multiplicity of the zero if $f(z_0) = 0$, and the negative of the order of the pole if $f$ has a pole in $z_0$].
Since the trivial zeros of $\zeta$ are all simple (which is easy to see from the functional equation), it follows that
$$\frac{\zeta'(s)}{\zeta(s)}$$
has simple poles with residue $1$ there, so
$$\operatorname{Res}\left(\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s}; -2n\right) = \frac{x^{-2n}}{-2n} = - \frac{1}{2n x^{2n}}$$
for $n$ a positive integer. Generally, if $\rho$ is a zero of $\zeta$ with multiplicity $\mu(\rho)$, we have
$$\operatorname{Res}\left(\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s};\rho\right) = \mu(\rho)\frac{x^\rho}{\rho}.$$
The pole of $\zeta$ at $1$ adds a simple pole of $\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s}$ with residue $-x$.
