Solving Integral Using Contours

Question

I am trying to evaluate the integral $$\int_{0}^{\infty}\frac{\sin^2(x)}{\sinh^2(x)}\;dx$$ by integrating $$f(z)=\frac{1-e^{2iz}}{\sinh^2(z)}$$ over a semicircular contour.

The horizontal segment of the contour returns back 4 times the integral we were looking for. In addition, on the part where we jump over the simple pole at $z=0$, we get a value of $-2\pi$. I checked that the infinite top arc goes to zero.

The residue calculation for the double poles at $z=in\pi$ for $n=1,2,...$ gives me an infinite geometric series. $$-2i\sum_{n=1}^{\infty}e^{-2n\pi}=-2i\frac{e^{-2\pi}}{1-e^{-2\pi}}$$.

All together, I get $$2\int_{0}^{\infty}\frac{1-\cos(2x)}{\sinh^2(x)}\;dx=4\int_{0}^{\infty}\frac{\sin^2(x)}{\sinh^2(x)}\;dx=2\pi+4\pi\frac{e^{-2\pi}}{1-e^{-2\pi}}$$.

But this is still way off from the correct answer of $\frac{1}{2}(-1+\pi\coth(\pi))$.

Maybe semicircular contours is not the ideal method. Rectangular contours results in divergences around the pole at $z=i\pi$. Please help.

Thank you.

I would argue why you would need to jump at $z=0$. The limit, in my knowledge, does exist. $\lim_{z\to0}\frac{\sin^2z}{\sinh^2z}=1$. So we don't need to jump at that point. — Laxmi Narayan Bhandari, Jul 08 '22 at 12:50
The function I am integrating on the contour is $f(z)$, as described above. That has a simple pole at $z=0$, according to my judgement. — Rudra Kamat, Jul 08 '22 at 12:59
We consider a point a pole only when the limit at the point tends to infinity, which isn't the case here. — Laxmi Narayan Bhandari, Jul 08 '22 at 13:01
We have $f(z)=\frac{1-e^{2iz}}{\sinh^2(z)}$. You can check with L'Hôpital's rule that the sinh square causes a simple pole at $z=0$. — Rudra Kamat, Jul 08 '22 at 13:04
Oops. I did a quick check and it seems like there actually is a pole at $z=0$. I'll try to do the integral myself, and answer if I find some mistake or anything else. — Laxmi Narayan Bhandari, Jul 08 '22 at 13:04
Note that your solution is equal to $2\pi\coth(\pi)$, so you're not as far off as you think. — eyeballfrog, Jul 08 '22 at 13:18
And see this this answer for the other approach you mentioned. — Random Variable, Jul 08 '22 at 13:48
Hey. Thanks a lot. I guess you get the result if you use an infinite rectangle contour as you have produced instead of a semicircle. Kind of weird tbh. — Rudra Kamat, Jul 08 '22 at 14:13
@RudraKamat The integral along the semicircle also goes to $1$ as $N \to \infty$. It's just easier to use an infinite rectangle. — Random Variable, Jul 08 '22 at 14:23

Solving Integral Using Contours

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