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?

  • $\begingroup$ Very tricky indeed. My computer tells me the answer is $\frac{1}{4}\pi\tau^2\omega\;\text{csch}(\frac{\pi\tau\omega}{2})$. Where csch is the hyperbolic cosecant: mathworld.wolfram.com/HyperbolicCosecant.html. I don't have an idea how to do this by hand. The indefinite integral is very ugly and involves the hypergeometric function. $\endgroup$ – Uncountable Feb 18 '15 at 22:25
  • $\begingroup$ Did you get this by Maple? $\endgroup$ – Libertron Feb 18 '15 at 22:26
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    $\begingroup$ I used Mathematica for the calculation (it is a similar program). $\endgroup$ – Uncountable Feb 18 '15 at 22:27
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    $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. In particular: why is this particular integral of interest? Where does it arise? $\endgroup$ – Carl Mummert Feb 20 '15 at 12:03

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


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


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.


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