I'm interested in an asymptotic expansion of the following Riemann zeta-type function
$$
\begin{align}
 \displaystyle \zeta(s \mid a,b) := \sum\_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}, 
 \quad \Re a >-1, \, \Re b >-1, \, s>0, \tag1
\end{align} 
$$ as $s \to 0^+$.


The case $a=b$ in $(1)$ leads to the Riemann Hurwitz zeta function with the Laurent expansion near $0$:  

$$
\begin{align} \sum\_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}}
= \frac{1}{s}+\sum\_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma\_{k}(a+1)s^{k}, \quad s>0, \tag2
\end{align}
$$ where $\displaystyle \gamma\_{k}(a+1)$ are the [generalized Stieltjes constants](http://en.wikipedia.org/wiki/Stieltjes_constants) with $\displaystyle \gamma\_{0}(a+1)=-\Gamma'(a+1)/\Gamma(a+1)$.


What is an asymptotic expansion, as $s \to 0^+$, of $\displaystyle \zeta(s \mid a,b)$ when $a\neq b$?



Asymptotic expansion of $\zeta(s \mid a,b)= \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}$
For $s > 0$ we have


$$
\sum\_{n=1}^{\infty} \frac{1}{(n+a)^s(n+b)} - \sum\_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \sum\_{n=1}^{\infty} \frac{a-b}{(n+a)^{s+1}(n+b)} =: g(s).
$$


The series on the right-hand side converges and is analytic on $\operatorname{Re} s > -1$, so the difference on the left-hand side can be analytically continued to this region. Consequently, the analytic continuation $\zeta(s \mid a,b)$ of your sum satisfies


$$
\zeta(s \mid a,b) = \zeta(s+1,a+1) + g(s)
$$


for $\operatorname{Re} s > -1$, $s \neq 0$ and thus has a simple pole at $s=0$ with residue $1$.


For $|s| < 1$ we have


$$
g(s) = (a-b)\sum\_{k=0}^\infty \left( \sum\_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!},
$$


so for $0 < |s| < 1$


$$
\zeta(s \mid a,b) = \frac{1}{s} + \sum\_{k=0}^{\infty} \left( \gamma\_{k}(a+1) + (a-b)\sum\_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}.
$$


