Complex integral using Residue Theorem with a regularised pole I need to prove the following integral computation by applying the residue theorem:
$$\int_{-\infty}^{+\infty}ds \ e^{-i\Omega s}\bigg(\frac{a^2}{4(\sinh{[\frac{a}{2}s]})^2}-\frac{1}{s^2}\bigg)=-2\pi\Omega\frac{1}{e^{\frac{2\pi\Omega}{a}}-1}$$
where $\Omega$ is a real positive number.
As far as I know, the integration countour could be chosen in the lower half plane ''as it is supposed that the pole in the origin doesn't exist''. But I don't understand that and thus I don't know how to do the integral 
I would appreciate some information and if it's possible the solution as well.
Thank you very much.
 A: I will assume that $a > 0$. If $a \in i\mathbb{R}$, the integral doesn't exist, since then the integrand has non-integrable poles (of order $2$) on the real line. If $a\in \mathbb{C}\setminus (\mathbb{R}\cup i\mathbb{R})$ the integral exists, but the evaluation is more complicated. For the remaining case $a\in \mathbb{R}\setminus \{0\}$, the invariance of the integrand under $a\mapsto -a$ allows to assume $a > 0$.
Since $\sinh z = z + O(z^3)$ near $0$, the principal part of $\dfrac{a^2}{4\sinh^2 \bigl(\frac{as}{2}\bigr)}$ at $0$ is $\dfrac{1}{s^2}$, hence
$$g(s) = \frac{a^2}{4\sinh^2\bigl(\frac{as}{2}\bigr)} - \frac{1}{s^2}$$
is holomorphic on a neighbourhood of $\mathbb{R}$.
Since $\Omega > 0$, and $\lvert e^{-i\Omega s}\rvert = e^{\operatorname{Re} - i\Omega s} = e^{\Omega\operatorname{Im} s}$, the exponential factor is small on the lower half-plane, and large on the upper. Therefore we cannot show that certain integrals tend to $0$ for contours in the upper half-plane with the standard estimate or similar tools, which is possible in the lower half-plane. Hence we choose paths in the lower half-plane to close our contour.
We don't have $\lim\limits_{s\to\infty} g(s) = 0$, however, $g$ has poles of order $2$ at $\frac{2\pi ik}{a}$ for $k\in \mathbb{Z}$, thus Jordan's lemma cannot be (directly) used, we need to take a different approach. To find appropriate contours, we must look at the behaviour of $\sinh$. We have
$$\lvert \sinh (x+iy)\rvert^2 = \lvert \sinh x \cos y + i\sin y \cosh x\rvert^2 = \sinh^2 x \cos^y + \cosh^2 x\sin^2 y = \sinh^2 x + \sin^2 y,$$
so on the lines $\operatorname{Re} s = \pm R$, we have
$$\lvert g(s)\rvert \leqslant \frac{a^2}{4\sinh^2\frac{aR}{2}} + \frac{1}{R^2}$$
and thus
$$\biggl\lvert \int_0^Y e^{-i\Omega(\pm R - it)} g(\pm R - it)\,dt\biggr\rvert \leqslant \biggl(\frac{a^2}{4\sinh^2\frac{aR}{2}} + \frac{1}{R^2}\biggr)\cdot \int_0^Y e^{-\Omega t}\,dt \leqslant \frac{1}{\Omega}\biggl(\frac{a^2}{4\sinh^2\frac{aR}{2}} + \frac{1}{R^2}\biggr).$$
The right hand side tends to $0$ as $R\to \infty$. If $y_k = \frac{(1-2k)\pi}{a}$, then we have $\lvert\sinh^2 \frac{as}{2}\rvert \geqslant \sin^2 \frac{ay_k}{2} = 1$ for $\operatorname{Im} s = y_k$, and therefore
$$\biggl\lvert \int_{-R}^R e^{-i\Omega(t+iy_k)} g(t+ iy_k)\,dt\biggr\rvert \leqslant \biggl(\frac{a^2}{4} + \frac{1}{y_k^2}\biggr)\cdot 2R\cdot e^{\Omega y_k}.$$
Given any fixed $R$, the right hand side tends to $0$ as $k \to +\infty$.
The integrand has poles in $z_n = \frac{-2\pi in}{a}$, $n\in \mathbb{Z}\setminus \{0\}$, and $y_k < \operatorname{Im} z_n < 0 \iff 0 < n \leqslant k$, so by the residue theorem, if $C_{R,k}$ denotes the rectangular contour with vertices $-R, R, R+ iy_k, -R+iy_k$, we have
$$\int_{C_{R,k}} e^{-i\Omega s}g(s)\,ds = -2\pi i \sum_{n = 1}^k \operatorname{Res}\bigl(e^{-i\Omega s} g(s); z_n\bigr).$$
Letting $R$ and $k$ tend to $+\infty$ in a suitable way (so that $R e^{\Omega y_k} \to 0$), we obtain
$$\int_{-\infty}^\infty e^{-i\Omega s} g(s)\,ds = - 2\pi i\sum_{n=1}^\infty \operatorname{Res}\bigl(e^{-i\Omega s} g(s); z_n\bigr).$$
It remains to find the residues. Due to the periodicity of $\sinh$, the coefficients of Laurent expansion of $\dfrac{a^2}{4\sinh^2 \frac{as}{2}}$ around $z_n$ are the same for all $n$, and hence
$$\operatorname{Res}\bigl(e^{-i\Omega s}g(s); z_n\bigr) = \operatorname{Res} \biggl(e^{-i\Omega s}\frac{a^2}{4\sinh^2\frac{as}{2}}; z_n\biggr) = e^{-i\Omega z_n}\operatorname{Res}\biggl(e^{-i\Omega s}\frac{a^2}{4\sinh^2\frac{as}{2}}; 0\biggr).$$
At $0$, we know the expansion is $e^{-i\Omega s} \bigl(\frac{1}{s^2} + h(s)\bigr)$ with a holomorphic $h$, so the residue there is $-i\Omega$ and
$$\int_{-\infty}^\infty e^{-i\Omega s} g(s)\,ds = -2\pi\Omega \sum_{n=1}^\infty e^{-i\Omega z_n} = -2\pi\Omega \frac{e^{-i\Omega z_1}}{1-e^{-i\Omega z_1}} = -\frac{2\pi\Omega}{e^{\frac{2\pi\Omega}{a}}-1},$$
since $e^{-i\Omega z_n} = \bigl(e^{-i\Omega z_1}\bigr)^n$ and $i\Omega z_1 = \frac{2\pi \Omega}{a}$.
