Complex integral computation with $\sinh$ I need to prove the following integral computation by applying the residue theorem: $$\int_{-\infty}^{+\infty}ds\frac{e^{-i\Omega s}}{(\sinh{[\frac{a}{2}s-i\epsilon]})^2}=-8\pi\frac{\Omega}{a^2}\frac{1}{e^{\frac{2\pi\Omega}{a}}-1}$$ As far as I know, there are many poles that make $0$ the denominator but the one that satisfies $\frac{a}{2}s-i\epsilon=0$ is avoided because the integration countour could be in the lower half plane thanks to the $-i\epsilon$ term. I would appreciate solution or hints. Thank you.

We assume that $\Omega$ is real-valued and positive.

The poles are located at $s_n = \frac{2i}{a}(n\pi+\epsilon)$. If we analyze the integrand, we see that

$$\frac{e^{-i\Omega s}}{\sinh^2(\frac{a}{2}(s-\frac{2i\epsilon}{a}))}=\frac{e^{-i\Omega s_n}\left(1-i\Omega(s-s_n)+O\left(s-s_n\right)^2\right) }{(a^2/4)(s-s_n)^2+O\left(s-s_n\right)^6}$$

which reveals that the residues are

$$\frac{4}{a^2}\left(-i\Omega\right)e^{-i\Omega s_n}=\frac{4}{a^2}\left(-i\Omega\right)e^{-\Omega(2\epsilon/a)}\left(e^{-\Omega (2\pi/a)}\right)^n$$

Summing these we have

$$\frac{4}{a^2}\left(-i\Omega\right)e^{-\Omega(2\epsilon/a)}\frac{1}{e^{\Omega (2\pi/a)}-1}$$

The value of the integral is $-2\pi i$ (since we close in the lower-half plane) times the sum of these residues and is

> $$-\frac{8\pi \Omega}{a^2}e^{-\Omega(2\epsilon/a)}\frac{1}{e^{\Omega (2\pi/a)}-1} \to -\frac{8\pi \Omega}{a^2}\frac{1}{e^{\Omega (2\pi/a)}-1} $$

as $\epsilon \to 0$ as expected!!