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

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?
 A: $$\color{blue}{\int_0^\infty {\frac{{\cos (\pi x)}}{{{e^{2\pi \sqrt x }} - 1}}dx} = \frac{{2 - \sqrt 2 }}{8}}$$
Simple manipulation gives
$$\int_0^\infty {\frac{{\cos (\pi x)}}{{{e^{2\pi \sqrt x }} - 1}}dx} = 2\int_0^\infty {\frac{{x\cos (\pi {x^2})}}{{{e^{2\pi x}} - 1}}dx} = \frac{1}{2} \int_0^\infty {\frac{{\sin (\pi {x^2})}}{{{{\sinh }^2}\pi x}}dx} $$
we evaluate the last integral.

Consider the function $$f(z) = \frac{{{e^{i\pi {z^2}}}{e^{2\pi z}}}}{{{{\sinh }^2}\pi z\sinh 2\pi z}}$$
note that $$f(z) - f(z + 2i) = 2\frac{{{e^{i\pi {z^2}}}}}{{{{\sinh }^2}\pi z}}$$
Integrate $f(z)$ around the rectangle with vertices $-R, R, R+2i, -R+2i$, with semicircle indentations at $0$ and $2i$. The indentation at $0$ is above the $x$-axis, while indentation at $2i$ is also above the line $\Im(z) = 2$, denote these two circles as $C_1$ and $C_2$ respectively, both with radius $r$. 
Then $$\tag{1} 2\pi i \sum_{k=1}^4 \text{Res}[f(z),\frac{k}{2}i] = 2 \mathcal{P}\mathcal{V} \int_{ - \infty }^\infty {\frac{{{e^{i\pi {z^2}}}}}{{{{\sinh }^2}\pi z}}dz} + \int_{C_1} f(z) dz + \int_{C_2} f(z) dz $$
As $r\to 0$, 
$$\begin{aligned}
\int_{C_1} f(z) dz + \int_{C_2} f(z) dz &= - \int_0^\pi {ri{e^{ix}}\left[ {f(r{e^{ix}}) - f(2i + r{e^{ix}})} \right]dx} \\
&= - \int_0^\pi {\frac{2}{{{\pi ^2}{r^2}{e^{2ix}}}}ri{e^{ix}}dx} + o(1) \\
&= - \frac{4}{{{\pi ^2}r}} + o(1)
\end{aligned}$$
where we used the expansion of $f(z)-f(z+2i)$ around $0$. Note the dominant term is real.
Thus $(1)$ becomes
$$2\pi i\left( {\frac{1}{\pi } - \frac{1}{{\sqrt 2 \pi }} + \frac{i}{{\sqrt 2 \pi }}} \right) = 2\mathcal{P}\mathcal{V}\int_{ - \infty }^\infty {\frac{{{e^{i\pi {z^2}}}}}{{{{\sinh }^2}\pi z}}dz} - \frac{4}{{{\pi ^2}r}} + o(1)$$
Comparing imaginary part gives
$$\int_0^\infty {\frac{{\sin (\pi {x^2})}}{{{{\sinh }^2}\pi x}}dx} = \frac{{2 - \sqrt 2 }}{4} $$

Denote $$I(a) = \int_0^\infty  {\frac{{\cos (a\pi x)}}{{{e^{2\pi \sqrt x }} - 1}}dx} $$
then additional values include

$$\begin{aligned}I(2) &= \frac{1}{{16}} \\ I(3) &= \frac{1}{4} - \frac{{\sqrt 2 }}{{24}} - \frac{{\sqrt 6 }}{{18}} \\ I(\frac{1}{2}) &= \frac{1}{{4\pi }} \\
I(\frac{1}{3}) &= 1 - \frac{{3\sqrt 6 }}{8} \\ I(\frac{2}{3}) &= \frac{1}{3} - \frac{{3\sqrt 3 }}{{16}} + \frac{{\sqrt 3 }}{{8\pi }} \end{aligned}$$

In general, $I(r)$ for rational $r$ is algebraic over $\mathbb{Q}(\pi)$.

I just realized the question has been answered here, but that answer is not quite complete.
