Continued fraction for some integrals by Ramanujan

Solution 1:

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 $$ 2\int_{0}^{\infty}\frac{x^{2}e^{-x\sqrt{3}}}{\sinh x}\,dx = \cfrac{1}{1+}\cfrac{1^{3}}{1+}\cfrac{1^{3}}{3+}\cfrac{2^{3}}{1+}\cfrac{2^{3}}{5+}\cfrac{3^{3}}{1+}\cfrac{3^{3}}{7+}\cdots $$