Integral of $\frac{1}{x-a}\frac{1}{e^{x-a}-1}$

Question

For positive powers of $x$, I know how to do this integral, but with $x$ at the denominator? Is there a way to at least isolate the divergence, or to do this using Cauchy's theorem? The integral is for $$|x^2-a^2|\leq 1$$ but I would like to know how to do it in several cases, indefinite and for $0$ to $\infty$. $x$ is real.

Answer 1

There is no closed-form antiderivative. Moreover the pole at $x=a$ has order $2$, so it is non-integrable.