I tried solving the integral $$ I := \int_{-\infty}^{\infty} \frac{x \exp(\mu x)}{\exp(\nu x)-1}\,dx,~~\text{where}~~\text{Re}(\nu)>\text{Re}(\mu) >0$$
using the calculus of residues. I found the first order poles and calculated the residues $$ z_n = \frac{2 n \pi i}{\nu},~~n \in \mathbb{N}\setminus \lbrace{ 0 \rbrace},~~\text{Res}_{z_n} f = \frac{2 n \pi i}{\nu^2} \exp\left(2 n \pi i \frac{\mu}{\nu} \right).$$ When I now sum over all residues to get the result for the integral I encounter a series $$ I = \dots = - \frac{4 \pi^2}{\nu^2} \sum_{n=1}^{\infty} n \exp\left(2 \pi i n \frac{\mu}{\nu} \right) = \dots = \left( \frac{\pi}{\nu} \csc\left( \frac{\pi \mu}{\nu} \right) \right)^2,$$ which I can evaluate by integrating the summands w.r.t $n$, then using the formula for the geometric series and differentiating w.r.t. $n$ again. This gives me the correct result (according to Mathematica). However to use the formula for the geometric series I have to assume that $$ \left|\exp\left( 2 \pi i \frac{\mu}{\nu} \right)\right| =\left|\exp\left( -2 \pi \text{Im}\left(\frac{\mu}{\nu} \right)\right)\right|<1$$ and that does not hold for all $\mu,\nu$ fulfilling $\text{Re}(\nu)>\text{Re}(\mu)>0$. Why does this approach give me the correct result although the series appears to diverge for some $\mu,\nu$ and is there some other way to solve this integral by hand?
You could split up the integral into two geometric series
$$\int_{-\infty}^\infty \frac{xe^{\mu x}dx}{e^{\nu x}-1} = \sum_{n=1}^\infty \int_0^\infty xe^{(\mu-n\nu)x}dx - \sum_{n=0}^\infty \int_{-\infty}^0 xe^{(\mu+n\nu)x}dx$$
$$ = \sum_{n=-\infty}^\infty \frac{1}{(\mu-n\nu)^2} = \frac{\pi^2}{\nu^2}\csc^2\left(\frac{\pi\mu}{\nu}\right)$$