Closed form for $\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$$
 A: 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
$$
I(a)= \frac{\pi}{\sqrt{2}}\left(\sin \left(\frac a 2 \right)J_0 \left(\frac a 2 \right)-\cos\left(\frac a 2 \right)Y_0\left(\frac a 2 \right) \right)
$$
A: Hint:
$\int_0^\infty\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}e^{-x}~dx$
$=\int_0^\infty\dfrac{\sqrt{\sinh x+\sqrt{\sinh^2x+1}}}{\sqrt{\sinh x}\sqrt{\sinh^2x+1}}e^{-\sinh x}~d(\sinh x)$
$=\int_0^\infty\dfrac{\sqrt{\sinh x+\cosh x}}{\sqrt{\sinh x}\cosh x}e^{-\sinh x}\cosh x~dx$
$=\int_0^\infty\dfrac{e^{-\sinh x}\sqrt{e^x}}{\sqrt{\dfrac{e^x-e^{-x}}{2}}}dx$
$=\sqrt2\int_0^\infty e^{-\sinh x}\sqrt{\dfrac{e^x}{e^x-e^{-x}}}~dx$
$=\sqrt2\int_0^\infty\dfrac{e^{x-\frac{e^x-e^{-x}}{2}}}{\sqrt{e^{2x}-1}}dx$
$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{e^x}{2}+\frac{1}{2e^x}}}{\sqrt{e^{2x}-1}}d(e^x)$
$=\sqrt2\int_1^\infty\dfrac{e^{-\frac{x}{2}+\frac{1}{2x}}}{\sqrt{x^2-1}}dx$
$=\sqrt2\int_1^\infty e^{-\frac{x}{2}+\frac{1}{2x}}~d(\cosh^{-1}x)$
$=\sqrt2\int_0^\infty e^{-\frac{\cosh x}{2}+\frac{1}{2\cosh x}}~dx$
$=\sqrt2\int_0^\infty e^\frac{1-\cosh^2x}{2\cosh x}~dx$
$=\sqrt2\int_0^\infty e^{-\frac{\sinh^2x}{2\cosh x}}~dx$
$=\sqrt2\int_0^\infty e^{-\frac{x^2}{2\sqrt{x^2+1}}}~d(\sinh^{-1}x)$
$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{x^2}{2\sqrt{x^2+1}}}}{\sqrt{x^2+1}}dx$
$=\sqrt2\int_\infty^0\dfrac{e^{-\frac{1}{2x^2\sqrt{\frac{1}{x^2}+1}}}}{\sqrt{\dfrac{1}{x^2}+1}}d\left(\dfrac{1}{x}\right)$
$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{1}{2x\sqrt{x^2+1}}}}{x\sqrt{x^2+1}}dx$
