<s>[INST] 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?
 [/INST] 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 $u^2=v u-1$.
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