TITLE: trouble with this integral
QUESTION [8 upvotes]: Could anyone help me to do this integral ?
$$\int_{\,0}^\infty \; \frac{\exp \left( -\frac{1}{x} -x\right)}{\sqrt{x}}  \, dx = \sqrt{\pi}e^{-2} $$
I think you start with completing the square in the exponent, but what substitution do you make then ? $u=\sqrt{x}$ didn't seem to get me far.

REPLY [10 votes]: Substitute first $x=u^2$ in order to have:
$$ I = \int_{0}^{+\infty}\frac{dx}{\sqrt{x}\exp\left(x+\frac{1}{x}\right)}=2\int_{0}^{+\infty}e^{-\left(x^2+\frac{1}{x^2}\right)}\,dx$$
Use now the substitution $x=\frac{1}{y}$ to have:
$$ I = 2\int_{0}^{+\infty}\frac{1}{x^2}e^{-\left(x^2+\frac{1}{x^2}\right)}\,dx,$$ 
from which follows:
$$ I = \int_{0}^{+\infty}\left(1+\frac{1}{x^2}\right)e^{-\left(x^2+\frac{1}{x^2}\right)}\,dx,$$ 
and the key substitution is now $u = x-\frac{1}{x}$, from which we have:
$$ I = \int_{-\infty}^{+\infty}e^{-u^2-2}\,du = e^{-2}\sqrt{\pi}, $$
QED.