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.

Solution 1:

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