Does $\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 16}?$ How do we prove these two results?
$$\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 16}\tag1$$
$$\int_{0}^{\infty}{\sin(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 8\pi}\tag2$$
Try an approach to split out the form 
$${\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 4}{\sin(16x^2\pi)-\sin(4x^2\pi)\over 2\sinh(x\pi)\cosh(x\pi)}\tag3$$
$$={1\over 4}{\sin(16x^2\pi)-4\sin(x^2\pi)\cos(x^2\pi)+8\sin^3(x^2\pi)\cos(x^2\pi)\over 2\sinh(x\pi)\cosh(x\pi)}\tag4$$
 Or we could write $(1)$ as 
$$\int_{0}^{\infty}{({e^{i10x^2\pi}+e^{-i10x^2\pi}})({e^{i6x^2\pi}-e^{-i6x^2\pi}})\over4i \sinh^2(2x\pi)}\mathrm dx\tag5$$
$$\int_{0}^{\infty}{{e^{i16x^2\pi}-e^{-i16x^2\pi}}-({e^{i4x^2\pi}-e^{-i4x^2\pi}})\over4i \sinh^2(2x\pi)}\mathrm dx\tag6$$
$$\int_{0}^{\infty}{\sinh(i16x^2\pi)-\sinh(i4x^2\pi)\over2i \sinh^2(2x\pi)}\mathrm dx\tag7$$
Surely this is not the correct  approach here.
I estimated the closed form using wolfram integrator, not sure it is correct.
 A: Maybe the most effective way to calculate these integrals is not to start from scratch and use residue theory, but to apply the general formula proved in Ramanujan's Lost Notebook, part IV, formula 14.4.14. In another words there is no need to reinvent the wheel. I will demonstrate the method and its effectiveness by calculating $(1)$. Note that
\begin{align}
I_1&=\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx\\
&=\int_{0}^{\infty}{\sin(16x^2\pi)-\sin(4x^2\pi)\over 2\sinh^2(2x\pi)}\mathrm dx\\
&=\int_{0}^{\infty}{32\pi x\cos(16x^2\pi)-8\pi x\cos(4x^2\pi)\over 4\pi}\left(\frac{1}{\tanh(2x\pi)}-1\right)\mathrm dx\\
&=\frac12\int_{0}^{\infty}{\left(4\cos(4x^2\pi)-\cos(x^2\pi)\right)}x\left(\frac{1}{\tanh(x\pi)}-1\right)~\mathrm dx
\end{align}
It is known that the following diverging integral can be regularized through the introduction of a regularization factor $e^{-\delta x^2},~\delta\to+0$
$$
\int_{0}^{\infty}x\cos\pi a x^2~dx=0\tag{*}.
$$
Ramanujan considers the function 
$$
F_w(t)=\int_{0}^{\infty}{\sin(\pi t x)\over\tanh(x\pi)}e^{-\pi wx^2}~\mathrm dx
$$
and proves the following proposition

where the prime $'$ on the summation sign indicates that the terms with $j = 0, n$ are to be multiplied by $\frac{1}{2}$.
This formula allows one to calculate $F_w(t)$ when $w=-\eta_2ni/(\eta_1 m)$, because in that case $s=t$ and 14.4.14 can be solved for $F_w(t)$.
For example, for odd $m$ and $n$ 14.4.14 becomes:
Now divide (14.5.3) by $t$ and take the limit $t\to 0$. Of course the integral will be divergent, however it is made convergent by substracting $4\int_{0}^{\infty}x\cos(\pi x^2)~dx=0$. For example for $m=n=1$ one gets
$$
4\int_{0}^{\infty}{x\cos(x^2\pi)\left(\frac{1}{\tanh(2x\pi)}-1\right)}~\mathrm dx=\frac{1}{2}+\frac{-1}{2}(-1)+\left(\frac{-1}{2}\cdot \frac{1}{\sqrt{2}}+\frac{1}{2}\frac{-1}{\sqrt{2}}\right)-0\\
=1-\frac{1}{\sqrt{2}}.\tag{**}
$$
Numerical check confirms that (**) is correct.
When one of $m,n$ is even, then analog of $14.5.3$ will have $\sinh(mt)$ on the LHS instead of $\cosh(mt)$, and in this case one needs to divide by $t^2$ and apply Lopital's rule to the RHS. Of course there is no need to calculate everything by hand, because formulas $14.4.14$ can be plugged in Mathematica and closed forms evaluated automatically, thus saving a great deal of time and effort.
A: I've actually no idea how to proof it, but the question got me interested and I tried to do a few things. You could try to break them all down to exponentials. This is what I wrote on MathType for formula (1):

I would suggest to substitute the $e^{2\pi x}$ into a value $a$ and eventually you should get four integrals with something like 
$ \left(\frac{ca^2}{a^2-1}\right) $. Evaluating this integral should give a logaritm and subsituting the value back could give the pretty number you were looking for. Good luck!
