TITLE: Closed form for $\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$
QUESTION [25 upvotes]: Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$$

REPLY [4 votes]: I introduce a parameter $a$,
$$
I(a)=\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-ax}dx
$$
then took a Mellin transform from $a$ to $s$,
$$
\mathcal{M}_{a\to s}[I(a)]=\Gamma(s)\int_0^\infty x^{-s-\frac{1}{2}}\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}\; dx
$$
which apparently equals
$$
\mathcal{M}_{a\to s}[I(a)]=\sqrt{2\pi} \frac{\Gamma(s)^2}{\Gamma(s+\frac{1}{2})}\cos\left(\frac{\pi s}{2}\right)\sec(\pi s)
$$
then the inverse Mellin transform gives
$$
I(a)= \frac{\pi}{\sqrt{2}}\left(\sin \left(\frac a 2 \right)J_0 \left(\frac a 2 \right)-\cos\left(\frac a 2 \right)Y_0\left(\frac a 2 \right) \right)
$$