How to evaluate $\int_{0}^{+\infty }\frac{x^{3}\sin\left ( (1/2)\pi x \right )}{e^{2\pi \sqrt{x}}-1}\mathrm{d}x$ I got an answer below

\begin{align*}
\int_{0}^{\infty }\frac{x^{3}\sin\left ( \frac{1}{2}\pi x \right )}{e^{2\pi \sqrt{x}}-1}\mathrm{d}x&=\frac{17}{16}-\frac{8}{3\pi }-\frac{7}{\pi ^{2}}+\frac{35}{2\pi ^{3}}-\frac{105}{16\pi ^{4}} \\&\approx 0.00145669538148559\cdots \cdots 
\end{align*}

which agree with mathematica.But I have no idea how to prove it.
 A: From integrals and series 'S':
Using the techniques in Ramanujan's
 paper "Some definite Integrals connected to Gauss's sums", I arrived at
$$\int_0^\infty \frac{\sin(tx)\sin\left(\dfrac{\pi x^2}{2} \right)}{\tanh(\pi x)}\, \mathrm{d}x=\frac{1}{\sqrt{2}}\sin\left(\frac{\pi}{4}+\frac{t^2}{2\pi} \right)\coth(t)-\frac{1}{2\sinh(t)}$$
My guess is that this equation can be manipulated to get the desired result. By manipulations I mean substitutions, differentiation, taking limits etc.
Here's the link to the paper:Some definite Integrals connected to Gauss's sums
A: 
So, I recently posted this 1st step to an answer. I thought that it was wrong, so I made it CW, but after further investigation, it actually was right. Here was that post, without CW

Consider the integral:
$$f(a)=\int_0^\infty \frac{\cos ax}{e^{2\pi\sqrt{x}}-1}\text{dx}$$
Therefore: $$I=f'''\left(\frac{\pi}{2}\right)$$
We can shift our integral to sum form, due to that $\cos$ term there.
$$f(a)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!} \int_0^\infty \frac{(ax)^{2n}}{e^{2\pi\sqrt{x}}-1}\text{dx}$$
Substituting $x\to x^2,\text{dx}\to2\text{dx}$
$$=2\sum_{n=0}^\infty\frac{(-1)^na^{2n}}{(2n)!} \int_0^\infty \frac{x^{4n+1}}{e^{2\pi x}-1}\text{dx}$$
According to W|A, that integral actually has a closed form! Don't ask me how it got that...
$$-\sum_{n=0}^\infty \frac{(-1)^na^{2n}}{(2n)!} (2\pi)^{-4n-2}\text{Li}_{4n+2}(1)\Gamma(4n+2)$$
This is only partial, but might help.
