calculus - Is there a rapider or more elegant way to evaluate $int_0^{+infty} frac{cos(pi x) text{d}x}{e^{2pi sqrt{x}}-1}$?

$$\int_0^{+\infty} \frac{\cos(\pi x)\ \text{d}x}{e^{2\pi \sqrt{x}}-1}$$

First attempt

• $x\to t^2$




• Geometric series by writing the denominator as $e^{2\pi t}(1 - e^{-2\pi t})$




• $\cos(\pi t^2) = \Re e^{i\pi t^2}$

This leads me to

$$2\sum_{k = 0}^{+\infty} \int_0^{+\infty} t e^{i\pi t^2}e^{-\alpha t}\ \text{d}t$$

Where $\alpha = 2\pi (k+1)$.

Now I thought about writing it again as

$$-2\sum_{k = 0}^{+\infty}\frac{d}{d\alpha} \int_0^{+\infty} e^{i\pi t^2}e^{-\alpha t}\ \text{d}t$$

The last integral can be evaluated with the use of the Imaginary Error Function, hence a Special Function method.

Yet it doesn't seem me the best way.

Second Attempt

Basically like the previous one with the difference that

• $\cos( \cdot )$ stays as it;




• $\pi t^2 \to z$;

And this brings

$$-\frac{1}{\sqrt{\pi}}\frac{d}{d\alpha} \sum_{k = 0}^{+\infty}\int_0^{+\infty} \frac{\cos(z)}{z} e^{-\alpha \sqrt{\frac{z}{\pi}}}\ \text{d}z$$

But in both cases what I am thinking are just numerical methods. Or at least I could give a try with the stationary phase but... meh.

I don't know if I can use residues for this, actually. Even if taking a look at the initial integral, there is this additional way:

$$\frac{1}{e^{2\pi t} -1} = \frac{1}{(e^{\pi t}+1)(e^{\pi t}-1)}$$

Which for example has a pole at $t = +i$...

But using residues I would obtain

$$\pi \cos(\pi)$$

Where as the correct numerical result (which I checked with Mathematica) is

$$\color{blue}{0.0732233(...)}$$

And it seems there is not a closed form for this.

Any hint/help?