I am looking at this integral $\displaystyle\int_{-\infty}^\infty dx \, \frac{e^{iax}}{\sinh^2{bx}}$.
Now dividing the integral into real and complex parts, respectively $\int_{-\infty}^{\infty} dx \frac{\cos{ax}}{\sinh^2{bx}}$ and $\int_{-\infty}^{\infty} dx \frac{\sin{ax}}{\sinh^2{bx}}$, one would think that because of the singularity at $x =0 $ both the integrals would diverge.
But this integral can be evaluated using contour integration and gives a finite answer: $\frac{a}{e^{2\pi a/b}-1}$.
(The integration can be performed by taking a rectangular contour - running along the real axis from minus to plus infinity - deformed near the singularity at zero to include it and then going up and running from plus to minus infinity parallel to the real axis (at $2\pi i /b$ 'distance' above it) - deformed near the singularity at $2\pi i /b$ to exclude it.
Now I am trying to get some insight into why this divergent looking integral does not diverge and how do we only get a real answer. Taylor expanding near $x=0$ suggests $1/x$ divergence for the real part and $\log x$ divergence for the imaginary part.
Any insight/ heuristic explanation as to why this naive expectation is not borne out will be greatly appreciated.
Update: Okay I figured it out. The place where I saw it done via the contour integral hadn't been explicit about it - the integral is actually divergent. In the contour integral as outlined above, one obtains a sum of 3 integrals $\displaystyle\int_{-\infty}^{-\epsilon} + \int_{\epsilon}^\infty +\oint_C $ where the last is an integral over the semicircular deformation of radius $\epsilon$. The finite answer given above is the sum of these three integrals.
This is quite like the standard $\displaystyle \int_{-\infty}^\infty dx\, \frac{\sin x}{x}$. There too using the method of contour integration obtains a finite answer as a sum of three integrals as above. However there one may show that the integral $\displaystyle \oint_C $ is finite for all $\epsilon$, so taking the limit of $\epsilon \rightarrow 0$ one can evaluate the real integral above exactly. Here however $\displaystyle \oint_C $ will be of $\mathcal{O}(1/\epsilon)$ and so the integral will diverge in the $\epsilon \rightarrow 0$ limit. So the finite answer above is not quite equal to the integral above, the divergence is hiding in the $\displaystyle \oint_C $ integral. Rather, it is what in physics parlance is called the 'finite part of the answer' or 'renormalized' solution.