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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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    @JoeKing: This is a special case of the integral dealt with [here](http://math.stackexchange.com/a/168515/26872). This is the case $n = t = 1/\sqrt{\pi}$.2012-10-15

1 Answers 1

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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.

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    Wow. I would **never** have worked that out. Was it obvious to you ?2012-10-15