TITLE: Proving that the given two integrals are equal
QUESTION [9 upvotes]: I am stuck up with this simple problem. 
If $\alpha \cdot \beta = \pi$, then show that $$\sqrt{\alpha}\int\limits_{0}^{\infty} \frac{e^{-x^{2}}}{\cosh{\alpha{x}}} \ \textrm{dx} = \sqrt{\beta} \int\limits_{0}^{\infty} \frac{e^{-x^{2}}}{\cosh{\beta{x}}} \ \textrm{dx}$$
I tried replacing $\cosh{x} = \frac{e^{x}+e^{-x}}{2}$ and tried doing some manipulations, but it's of no use. Seems to be a clever problem. Moreover since we have $\alpha \cdot \beta = \pi$, we get $\sqrt{\alpha} = \frac{\sqrt{\pi}}{\sqrt{\beta}}$, but the Beta factor is in the numerator, which bewilders me.

REPLY [5 votes]: There is a very simple proof, based on some common facts.
Consider the self-reciprocal Fourier transformations
$$
\sqrt{\frac{2}{\pi}}\int_0^\infty e^{-x^2/2}\cos ax dx=e^{-a^2/2}
$$
$$
\sqrt{\frac{2}{\pi}}\int_0^\infty \frac{\cos ax}{\cosh \sqrt{\frac{\pi}{2}} x}dx=\frac{1}{\cosh \sqrt{\frac{\pi}{2}}a}.
$$
For two arbitrary self-reciprocal functions $f$ and $g$ we have
$$
\int_0^\infty f(x)g(\alpha x) dx=\int_0^\infty f(x)dx\cdot \sqrt{\frac{2}{\pi}}\int_0^\infty g(y) \cos(\alpha x y)dy=\\
\int_0^\infty g(y)dy \cdot \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos(\alpha x y)dx=\int_0^\infty g(y)f(\alpha y) dy=\\
\frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.
$$
Substituting the above two functions into this general formula we get
$$
\int\limits_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\alpha{x}}} {dx} = \frac{1}{\alpha} \int\limits_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\frac{x}{\alpha}}}{dx},
$$
which by trivial arithmetics can be brought to the symmetrical form given by Hardy.
For more information about self-reciprocal functions see my post here.