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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.

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Are you familiar with Error Functions? – Inquest Oct 15 '12 at 18:17
Does $\displaystyle \frac{1}{\sqrt x}= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-xt^2} dt$ help? Anyway, (+1) for the question. – user 1618033 Oct 15 '12 at 18:18
@JoeKing: This is a special case of the integral dealt with here. This is the case $n = t = 1/\sqrt{\pi}$. – user26872 Oct 15 '12 at 20:17
up vote 10 down vote accepted

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:

$$ I = \int_{-\infty}^{+\infty}e^{-u^2-2}\,du = e^{-2}\sqrt{\pi}, $$


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Nice and simple (+1). – user 1618033 Oct 15 '12 at 18:30
Wow. I would never have worked that out. Was it obvious to you ? – Joe King Oct 15 '12 at 18:45

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