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