Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.



> 
> **Question.** Is this true? If so, is there an underlying transformation or just a proof?
> $$\int\_0^{\infty}x^2e^{-x^2}\frac{{dx}}{\cosh\sqrt{\pi}x}
> =\frac14\int\_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$
> 
> 
> 



Is there a transformation or a proof for these integrals?
This is a generous explanation of Lucia's comment above.


The functions
$$\mathcal H\_n(x)=\frac{2^{1/4}}{(2^n n!)^{1/2}}H\_n(\sqrt{2\pi}\; x) e^{-\pi x^2}$$
form an orthonormal system in $L^2(\textbf{R})$. Here $H\_n(x)$ are the usual Hermite polynomials
defined by 
$$e^{2xz-z^2}=\sum\_{n=0}^\infty \frac{H\_n(x)}{n!}z^n,\qquad |z|<\infty.$$
The functions are eigenfunctions for the usual Fourier transform
so that 
$$\int\_{-\infty}^{+\infty}\mathcal H\_n(t)e^{-2\pi i x t}\,dt=(-i)^n \mathcal H\_n(x).$$


It follows that $L^2(\textbf{R})$ is a direct sum of four subspaces. In each of these subspaces
the Fourier transform is just multiplication by $1$, $i$, $-1$, $-1$ respectively.


It is well known that the function $1/\cosh\pi x$ is invariant by the Fourier transform
$$\int\_{-\infty}^{+\infty}\frac{e^{-2\pi i x\xi}}{\cosh\pi x}\,dx=
\frac{1}{\cosh\pi \xi}.$$
Therefore this function is in the span of the functions $\mathcal H\_{4n}(x)$, and 
therefore is orthogonal to any other eigenfunction. In particular 
$$\int\_{-\infty}^{+\infty}\frac{\mathcal H\_2(x)}{\cosh\pi x}\,dx=0.$$
Since $H\_2(x)=-2+4x^2$ it follows that 
$$\int\_{-\infty}^{+\infty}\frac{(8\pi x^2-2)e^{-\pi x^2}}{\cosh\pi x}\,dx=0.$$
Putting $x/\sqrt{\pi}$ instead of x 
$$\int\_{-\infty}^{+\infty}\frac{(8 x^2-2)e^{-x^2}}{\cosh\sqrt{\pi} x}\,\frac{dx}{\sqrt{\pi}}=0.$$
This is equivalent to your equation.


