Tricky integral with infinite limits

Question

I've been trying for some time in solving the following infinite integral. Will residue theory be of any help here? I haven't tried that yet, but it seems no method is working effectively. I want to evaluate $$ \int_{-\infty}^{\infty} \frac{e^{i \omega t}e^{-2t/ \tau}}{(1+ e^{-2t/ \tau})^2} dt \text{,}$$ where $\omega,\tau$ are arbitrary constants. Any useful trigonometric methods?

Answer 1

Let's choose $\tau,\omega>0$, the other cases can be done accordingly. I denote the integrand by $J(\tau)$ and the integral by $I(\tau)$

1.)

The structure of the integrand is as follows:

-it converges in the upper half of the complex plane

-changing $t\rightarrow t+i\pi \tau$ only results in a phase: $J(\tau)\rightarrow e^{-\pi \omega \tau}J(\tau)$

-it has poles at $t= (2N+1)\pi \tau /2 $ where $N$ is an integer number

2.)

The observations in 1.) suggest that we try to solve the problem by Contour integration using a rectangle with vertices $\{-\infty,0\},\{-\infty,i\pi \tau\},\{\infty,0\},\{\infty,i\pi \tau\}$ . The vertical parts vanish, and we end up with

$$ I(\tau)-e^{-\pi \omega \tau}I(\tau)=2 i \pi \text{Res}(z=e^{i\pi\tau\omega/2})\\\rightarrow I(\tau)=2 i\pi \times \frac{1}{1-e^{-\pi \omega \tau}} \times \left( -\frac{1}{4}i \omega \tau^2 e^{-\pi\omega \tau/2}\right)=\frac{\pi}{4}\frac{\omega \tau^2}{\sinh(\pi\omega\tau/2)} $$

Note: One could also use a big semicircle in the UHP and sum up all the residues.