Continued fraction for some integrals by Ramanujan While browsing through some letters of Ramanujan to G. H. Hardy I came across these two formulas (which are as mysterious as any other formula of Ramanujan): $$4\int_{0}^{\infty}\frac{xe^{-x\sqrt{5}}}{\cosh x}\,dx = \cfrac{1}{1+}\cfrac{1^{2}}{1+}\cfrac{1^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{3^{2}}{1+}\cfrac{3^{2}}{1+}\cdots\tag{1}$$ $$2\int_{0}^{\infty}\frac{x^{2}e^{-x\sqrt{3}}}{\sinh x}\,dx = \cfrac{1}{1+}\cfrac{1^{3}}{1+}\cfrac{1^{3}}{3+}\cfrac{2^{3}}{1+}\cfrac{2^{3}}{5+}\cfrac{3^{3}}{1+}\cfrac{3^{3}}{7+}\cdots\tag{2}$$ From these it appears that Ramanujan had general formulas for the integrals $$I_{m, n} = \int_{0}^{\infty}\frac{x^{m}e^{-x\sqrt{n}}}{\cosh x}\,dx,\,J_{m, n} = \int_{0}^{\infty}\frac{x^{m}e^{-x\sqrt{n}}}{\sinh x}\,dx\tag{3}$$ and for some values of $m, n$ he was able to express these as continued fractions.

I am totally perplexed by these formulas. Do we have any proofs of these or perhaps any reference which treats the integrals $I_{m, n}, J_{m, n}$?


Just for clarity the notation $$\cfrac{a_{1}}{b_{1}+}\cfrac{a_{2}}{b_{2}+}\cfrac{a_{3}}{b_{3}+}\cdots$$ consumes less space in typing compared to the more cumbersome but easier to understand continued fraction $$\cfrac{a_{1}}{b_{1} + \cfrac{a_{2}}{b_{2} + \cfrac{a_{3}}{b_{3} + \cdots}}}$$
 A: The following continued fraction is given on page 151 of Berndt's book "Ramanujan's Notebooks", Vol. II:

Let $x=\sqrt{5}$, and $n\to 0$ to obtain
\begin{align}
\int_{0}^{\infty}e^{-u\sqrt{5}}\frac{u}{\cosh u}\,du &= \cfrac{1}{4+}\cfrac{2^{2}}{1+}\cfrac{2^{2}}{4+}\cfrac{4^{2}}{1+}\cfrac{4^{2}}{4+}\cfrac{6^{2}}{1+}\cfrac{6^{2}}{4+}\ldots\\
&=\frac14 \cdot \cfrac{1}{1+}\cfrac{1^{2}}{1+}\cfrac{1^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{2^{2}}{1+}\cfrac{3^{2}}{1+}\cfrac{3^{2}}{1+}\ldots
\end{align}

$(2)$ is obtained from the continued fraction (32.4) on page 153
$\tag{32.4}$
Note that 
\begin{align}
2\int_{0}^{\infty}\frac{u^{2}e^{-u\sqrt{3}}}{\sinh u}\,du&=\sum_{n=0}^\infty 4\int_{0}^{\infty}u^{2}e^{-u\sqrt{3}-u-2nu}\,du\\
&=8\sum_{n=0}^\infty\frac{1}{(2n+1+\sqrt3)^3}\\
&=\sum_{n=1}^\infty\frac{1}{\left(n+\tfrac{\sqrt3-1}{2}\right)^3}.
\end{align}
Therefore (32.4) with $x=\frac{\sqrt3-1}{2}$ (i.e. $2x(x+1)=1$) gives
$$
2\int_{0}^{\infty}\frac{x^{2}e^{-x\sqrt{3}}}{\sinh x}\,dx = \cfrac{1}{1+}\cfrac{1^{3}}{1+}\cfrac{1^{3}}{3+}\cfrac{2^{3}}{1+}\cfrac{2^{3}}{5+}\cfrac{3^{3}}{1+}\cfrac{3^{3}}{7+}\cdots
$$
