AESTHETICALLY PLEASING INTEGRALS

I have decided to start a subsection (subset) of this blog by talking, and solving, the most beautiful (yes, aesthetically pleasant) integrals I have come across to during my "having fun time" periods, or generally during my studying periods. There are some of the which are really amazing, and they are amazing also because me myself and I have solved them. Sure, I don't claim to be the first one who solved them, since they are nothing special. But they are "hard enough" to make me compliment to myself for my courage and my will to attack them :)

THE ZEROTH API (Aesthetically Pleasing Integral)

$$J = \int_0^{+\infty} \dfrac{\sin(\pi x^2)}{\sinh^2(\pi x)}\ \text{d}x$$

Whose numerical result is $\dfrac{2 - \sqrt{2}}{4}$

PROOF

First of all we notice the integrand is even, hence we can write

$$\dfrac{1}{2}\int_{-\infty}^{+\infty}\dfrac{\sin(\pi x^2)}{\sinh^2(\pi x)}\ \text{d}x$$

Let's now define

$$f(z) = \dfrac{\cos(\pi z^2)}{\sinh(2\pi z) \sinh^2(\pi x)}$$

And not that because 

$$f(x\pm i) = \dfrac{-\cos(\pi x^2)\cosh(2\pi x) \pm i\sin(\pi x^2)\sinh(2\pi x)}{\sinh(2\pi x)\sinh^2(\pi x)}$$

we have

$$\int_{\gamma} f(z)\ \text{d}z = \int_{-\infty}^{+\infty} \left(f(x-i) + f(x+i)\right)\ \text{d}x$$

That is $2\pi i \times \text{Sum of the residues}$, which becomes $-\dfrac{\pi}{2} \times$ Sum of residues.

The poles are at $0$ and $\pm \dfrac{i}{2}$ and $\pm i$ hence we have: 

RESIDUES

If you are able enough to calculate the residues, which are rather easy, you'll find that the residue near zero is $-\dfrac{1}{2\pi}$

Instead we have $\dfrac{\sqrt{2}}{4\pi}$ near $\pm \dfrac{i}{2}$, whereas the residues around $\pm i$ are $-\dfrac{1}{2\pi}$.

Hence

$$\int_0^{+\infty} \dfrac{\sin(\pi x^2)}{\sinh^2(\pi x)}\ \text{d}x = \dfrac{\pi}{2}\left(-\dfrac{1}{2\pi} - \dfrac{1}{2\pi} + \dfrac{\sqrt{2}}{4\pi} + \dfrac{\sqrt{2}}{4\pi}\right) = \dfrac{2 - \sqrt{2}}{4}$$

As wanted.