Ramanujan Integral Identity assuming $\alpha\beta=\pi^2$

Question

In The Man Who Knew Infinity, Kanigel discusses Ramanujan's original letter to Hardy. In it, Ramanujan states that if $\alpha\beta=\pi^2$, then $$ \alpha^{-1/4}\left(1+4\alpha\int_0^\infty\frac{xe^{-\alpha x^2}}{e^{2\pi x}-1} \, dx \right)=\beta^{-1/4}\left(1+4\beta\int_0^\infty\frac{xe^{-\beta x^2}}{e^{2\pi x}-1} \, dx\right). $$ Does anyone know of a reference to a proof of this fact? There is no evidence offered in the book, other than a "Hardy had proved theorems like it, had even offered a similar one as a mathematical question...years before."

I am pretty unfamiliar with Ramanujan's work, so I don't really know where a good starting point would be for such a problem. Possibly using contour integration to turn it into an infinite sum via residues? I am a bit skeptical of this, as I have heard that Ramanujan was not very familiar with complex analysis at all.

I'm also not sure what the best way to search for this would be, despite looking for "Ramanujan integral identities," of which there are many to search through...

Thanks in advance for any help.

Answer 1

Start with the Bernoulli generating function $$\sum_{n=0}^{\infty} \frac{B_{n} \, x^n}{n!} = \frac{x}{e^{x} - 1}$$ to obtain \begin{align} I &= \int_{0}^{\infty} \frac{x \, e^{-\alpha \, x^{2}}}{e^{2\pi x} - 1} \, dx \\ 4 \pi^{2} \, I &= \int_{0}^{\infty} \frac{u \, e^{- p \, u^2}}{e^{u} - 1} \, dx \hspace{5mm} \text{ $u = 2\pi x$ and $p = \frac{\alpha}{4 \pi^{2}}$} \\ &= \sum_{n=0}^{\infty} \frac{B_{n}}{n!} \, \int_{0}^{\infty} e^{- p \, u^2} \, u^{n} \, dx \\ &= \frac{1}{2 \sqrt{p}} \, \sum_{n=0}^{\infty} \frac{B_{n}}{n!} \, \left(\frac{1}{p}\right)^{n/2} \, \Gamma\left(\frac{n+1}{2}\right) \\ &= \frac{1}{2 \sqrt{p}} \, \left[ - \frac{1}{2 \sqrt{p}} + \sqrt{\pi} \, \sum_{n=0}^{\infty} \frac{B_{2n}}{n!} \, \left(\frac{1}{4 p}\right)^{n} \right] \\ &= - \frac{\pi^{2}}{\alpha} + \pi \, \sqrt{\frac{\pi}{\alpha}} \, \sum_{n=0}^{\infty} \frac{B_{2n}}{n!} \, \left(\frac{\pi^{2}}{\alpha}\right)^{n} \end{align}

This leads to \begin{align} \alpha^{-1/4}\left(1+4\alpha\int_0^\infty\frac{xe^{-\alpha x^2}}{e^{2\pi x}-1}dx\right) = \frac{\alpha^{1/4}}{\sqrt{\pi}} \, \sum_{n=0}^{\infty} \frac{B_{2n}}{n!} \, \left(\frac{\pi^{2}}{\alpha}\right)^{n}. \end{align}

From here it is just making the connection.

Answer 2

I believe Ramanujan used self-Fourier transforms to prove this, as he did for another similar formula in his first letter. Consider the two sine transforms $$ \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) \sin(2\pi xt) \, dt = \frac{1}{2} \left(\frac{1}{e^{2\pi x}-1} - \frac{1}{2\pi x} \right), \hspace{0.5cm} x>0 $$ and $$ \int_0^{\infty} te^{-\pi \alpha t^2} \sin(2\pi xt) \, dt = \frac{1}{2} \alpha^{-3/2} xe^{-\pi x^2/\alpha}, \hspace{0.5cm} x,\; \text{Re }\alpha >0. $$ Applying the Fubini theorem we have $$ \int_0^{\infty} \left(\frac{1}{e^{2\pi x}-1}-\frac{1}{2\pi x} \right) xe^{-\pi \alpha x^2} \, dx = \int_0^{\infty} 2x e^{-\pi \alpha x^2} \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) \sin(2\pi xt) \, dt \, dx $$ $$ = \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) \int_0^{\infty} 2x e^{-\pi \alpha x^2} \sin(2\pi xt) \, dx \, dt = \alpha^{-3/2} \int_0^{\infty} \left(\frac{1}{e^{2\pi t}-1}-\frac{1}{2\pi t} \right) t e^{-\pi t^2/\alpha} \, dt $$ and therefore $$ \int_0^{\infty} \frac{xe^{-\pi \alpha x^2}}{e^{2\pi x}-1} \, dx - \frac{1}{4\pi \alpha^{1/2}} = \alpha^{-3/2} \int_0^{\infty} \frac{te^{-\pi t^2/\alpha}}{e^{2\pi t}-1} \, dt - \frac{1}{4\pi \alpha}, \hspace{0.5cm} \text{Re } \alpha>0 $$ where we have evaluated the two Gaussian integrals. Replacing $\pi \alpha$ with $\alpha$ and $\pi/\alpha$ with $\beta$ gives Ramanujan's symmetric expression.