Proving that the given two integrals are equal 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.
 A: 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.
A: This was shown by Hardy back in 1903/1904.
A mention of it can be found here: Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203, which is somewhere in the middle of a long paper.
Here is a snapshot in case that link does not work:

Note, the integral is slightly different, but I suppose it won't be too hard to convert it into the form you have.
See also Hardy's response to Ramanujan here: http://books.google.com/books?id=Of5G0r6DQiEC&pg=PA46. Note: 1b.

(Edit:) Since the journal itself has no reliable electronic copies, and the proof is actually somewhat more involved then just the excerpt shown above, I'll give a quick description of the proof that Hardy provided. 


*

*First is the concept of reciprocal functions of the first and second kind introduced by Cauchy. Two functions $\phi$ and $\psi$ defined on the positive real line is called reciprocal functions of the first kind if $$\phi(y) = \sqrt{\frac{2}{\pi}} \int_0^\infty \cos(y x) \psi(x) dx$$ and also the same formula with $\phi$ and $\psi$ swapped. They are called reciprocal functions of the second kind if the $\cos$ in the formula above is replaced by $\sin$. Cauchy gave several examples of each, and also examples of functions which are their own reciprocal function of the first kind (but not for the second), and proved that those functions have the following property: whenever $\alpha \beta = \pi$ $$\sqrt\alpha \left( \frac12 \phi(0) + \phi(\alpha) + \phi(2\alpha) + \cdots\right) = \sqrt\beta \left(\frac12 \psi(0) + \psi(\beta) + \psi(2\beta) + \cdots \right)$$

*In the article linked above, Hardy proved the following two facts (among others).


*

*The function $f(x) = e^{x^2/2}\int_x^\infty e^{-t^2/2}dt$ is its own reciprocal function of the second kind. (That proof is about 3 pages long, condensed in the typical Hardy fashion.)

*If $\phi$ and $\psi$ are reciprocal functions of the second kind, the following summation formula (analogue of the one above for functions of the first kind) holds: when $\lambda \mu = 2\pi$, one has $$ \sqrt\lambda \sum_0^\infty (-1)^n \phi\left( (n + \frac12)\lambda\right) = \sqrt\mu \sum_0^\infty (-1)^n \psi\left( (n+\frac12)\mu\right)$$ This expression being the one termed equation (9) in the screenshot above. 


*Hardy provided two proves of the formula asked about above in the question. The first proof proceeds by giving the series expansion $$\int_0^\infty \frac{e^{-\alpha x^2}}{\cosh \pi x} dx = \frac{2}{\pi} \sum (-1)^n F\left( (n + \frac12)\alpha\right)$$ where $$F(x) = \sqrt\pi e^{x^2}\int_x^\infty e^{-t^2}dt$$ and using equation (9) above. The second proof is shown in section 10 in the image above: he obtained a different series expansion of the expression we want on the left hand side, which can be shown to be term by term equal to the first series expansion of the expression on the right hand side, avoiding the need to invoke equation (9). 
