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

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.

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 $\alpha/\pi$ with $\beta$ gives Ramanujan's symmetric expression.