How to compute $\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$? How to compute this integral? :
$$\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$$
Wolframalpha gives the answer $\dfrac{3\sqrt{\pi}}{2e^2}$, but how to compute this?
 A: A little roundabout, but here goes.  Write
$$\begin{align}I &= \underbrace{\int_0^{\infty} dx \, \sqrt{x} \, e^{-\left (x+\frac1{x} \right )}}_{x=u^2} \\ &= 2 e^2 \underbrace{\int_0^{\infty} du \, u^2 \, e^{-\left (u+\frac1{u} \right )^2}}_{v=u+\frac1{u}}\\ &= e^2 \int_{\infty}^2 dv \, \left (1-\frac{v}{\sqrt{v^2-4}} \right )\left (\frac{v^2}{2}-\frac{v}{2} \sqrt{v^2-4}-1 \right )\, e^{-v^2}\\&+e^2 \int_2^{\infty} dv \, \left (1+\frac{v}{\sqrt{v^2-4}} \right )\left (\frac{v^2}{2}+\frac{v}{2} \sqrt{v^2-4}-1 \right )\, e^{-v^2}\\ &= 2 e^2 \underbrace{\int_2^{\infty} dv \,v \left ( \sqrt{v^2-4}+\frac1{\sqrt{v^2-4}}\right )\, e^{-v^2}}_{y=v^2-4}\\ &= e^{-2} \int_0^{\infty} dy \, \left (\sqrt{y}+\frac1{\sqrt{y}} \right ) e^{-y}\\ &= e^{-2} \left (\frac{\sqrt{\pi}}{2} + \sqrt{\pi} \right ) \\ &= \frac{3 \sqrt{\pi}}{2 e^2}\end{align}$$
For a little more background on how the integral over $v$ gets split, see this answer.  Also note that, in the third step, I made use of the fact that $u^2=v u-1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\root{x}
\exp\pars{-x - {1 \over x}}\,\dd x}
\\[5mm] \stackrel{x\ =\ \expo{2\theta}}{=}\,\,\,&
\int_{-\infty}^{\infty}\expo{\theta}
\exp\pars{-2\cosh\pars{2\theta}}\pars{2\expo{2\theta}}\,\dd\theta
\\[5mm] = &\
4\int_{0}^{\infty}\cosh\pars{3\theta}
\exp\pars{-2\cosh\pars{2\theta}}\,\dd\theta
\\[5mm] = &\
4\int_{0}^{\infty}\
\overbrace{\cosh\pars{\theta}\bracks{4\sinh^{2}\pars{\theta} + 1}}^{\ds{\cosh\pars{3\theta}}}\
\exp\pars{-4\sinh^{2}\pars{\theta} - 2}\,\dd\theta
\\[5mm] \stackrel{t\ =\ \sinh{\theta}}{=}\,\,\,\,\,\,&
4\expo{-2}\int_{0}^{\infty}\pars{4t^{2} + 1}\expo{-4t^{2}}\,\dd t =
\bbx{3\root{\pi} \over 2\expo{2}} \approx 0.3598 \\ &
\end{align}
A: With $t=\sqrt x$
\begin{align}
\int_0^{\infty}\sqrt x e^{-x-\frac{1}{x}} \, dx
&= \int_{-\infty}^{\infty}t^2e^{-t^2-\frac1{t^2}}dt
= \frac1{e^2}\int_{-\infty}^{\infty}t^2e^{-(t-\frac1{t})^2}dt\\
&= \frac1{e^2}\left(\int_{-\infty}^{\infty}\left(t^2 -1\right)e^{-(t-\frac1{t})^2}dt +\int_{-\infty}^{\infty}\underset{t\to \frac1t}{e^{-(t-\frac1{t})^2}}dt\right)\\
&=\frac1{e^2}\int_{-\infty}^{\infty}\left((t-\frac1t)^2 +1 \right)e^{-(t-\frac1{t})^2}dt \\
&= \frac1{e^2}\int_{-\infty}^{\infty}\left(t^2 +1 \right)e^{-t^2}dt 
=\frac{3\sqrt\pi}{2e^2}\\
\end{align}
where $\int_{-\infty}^{\infty}f(t-\frac1t)dt = \int_{-\infty}^{\infty}f(t)dt$ is used.
