<s>[INST] tough integral involving $\sin(x^2)$ and $\sinh^2 (x)$ I ran across this integral I get no where with. Can someone suggest a method of attack?.
$$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$
I tried series, imaginary parts, and so forth, but have made no progress.  
Thanks very much. 
 [/INST] Although this question is two years old, the integral was mentioned in chat recently, I evaluated it, and then found this question. Since there is no complete solution, although Hans Lundmark's suggestion is excellent and similar in nature, I am posting what I have done.
Contours
Since the integrand is even,
$$
\begin{align}
\int_0^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
\end{align}
$$
Define
$$
f(z)=\frac{\cos\left(\pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}
$$
Note that because
$$
f(x\pm i)
=\frac{-\cos\left(\pi x^2\right)\cosh(2\pi x)\pm i\sin\left(\pi x^2\right)\sinh(2\pi x)}{\sinh(2\pi x)\sinh^2(\pi x)}\\
$$
we have
$$
\begin{align}
\int_\gamma f(z)\,\mathrm{d}z
&=\int_{-\infty}^\infty\big[f(x-i)-f(x+i)\big]\,\mathrm{d}x\\
&=-2i\int_{-\infty}^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x\\
&=2\pi i\times\begin{array}{}\text{the sum of the residues}\\\text{inside the contour}\end{array}
\end{align}
$$
where $\gamma$ is the contour
$\hspace{3.2cm}$
Therefore,
$$
\int_0^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
=-\frac\pi2\times\begin{array}{}\text{the sum of the residues}\\\text{inside the contour}\end{array}
$$
Residues
near $0$ :
$$
\begin{align}
f(z)
&=\frac{\cos\left(\pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}\\
&=\frac{1-\frac12\pi^2z^4+O(z^8)}{2\pi z\left(1+\frac23\pi^2z^2+O(z^4)\right)\pi^2 z^2\left(1+\frac13\pi^2z^2+O(z^4)\right)}\\
&=\frac{1-\pi^2z^2}{2\pi^3z^3}+O(z)\\[10pt]
&\implies\text{residue}=-\frac1{2\pi}
\end{align}
$$
at $\pm i/2$, use L'Hosptal :
$$
\begin{align}
\text{residue}
&=\lim_{z\to\pm i/2}\frac{(z\mp i/2)\cos\left(\pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}\\
&=\frac1{2\pi\cosh(\pm\pi i)}\frac{\cos(-\pi/4)}{\sinh^2(\pm\pi i/2)}\\
&=\frac1{2\pi\cos(\pm\pi)}\frac{\sqrt2/2}{-\sin^2(\pm\pi/2)}\\[4pt]
&=\frac{\sqrt2}{4\pi}
\end{align}
$$
near $\pm i$ :
$$
\begin{align}
f(z\pm i)
&=\frac{-\cos\left(\pi z^2\right)\cosh(2\pi z)\pm i\sin\left(\pi z^2\right)\sinh(2\pi z)}{\sinh(2\pi z)\sinh^2(\pi z)}\\
&=\frac{-\left(1-\frac12\pi^2z^4+O(z^8)\right)\left(1+2\pi^2z^2+O(z^4)\right)+O(z^3)}{2\pi z\left(1+\frac23\pi^2z^2+O(z^4)\right)\pi^2 z^2\left(1+\frac13\pi^2z^2+O(z^4)\right)}\\
&=-\frac{1+\pi^2z^2}{2\pi^3z^3}+O(1)\\[10pt]
&\implies\text{residue}=-\frac1{2\pi}
\end{align}
$$
Result
Thus,
$$
\begin{align}
\int_0^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
&=-\frac\pi2\left(-\frac1{2\pi}-\frac1{2\pi}+\frac{\sqrt2}{4\pi}+\frac{\sqrt2}{4\pi}\right)\\[6pt]
&=\frac{2-\sqrt2}{4}
\end{align}
$$
</s>