Evaluating $\int_0^\infty e^{-(x-\frac{1}{x})^2} dx$ $$\int_0^\infty e^{-(x-\frac{1}{x})^2} dx$$
Any hints?
 A: Hint. One may apply the following result, 
$$
\int_{-\infty}^{+\infty}f\left(x-\frac{a}x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x  ,\qquad a>0,
$$ which is true for any integrable $f$  over $\mathbb{R}$.
A: Putting $u=x-1/x$, $x = \frac{1}{2}u+\sqrt{4+u^2} $ so $dx = \left( 1 + \frac{u}{\sqrt{u^2+4}} \right)\frac{du}{2} $. The interval of integration changes to $(-\infty,\infty)$, so you have
$$ \int_{-\infty}^{\infty} e^{-u^2}\left( 1 + \frac{u}{\sqrt{u^2+4}} \right)\frac{du}{2} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-u^2} \, du $$
since the second term is odd and hence has integral zero. Of course, we know $\int_{-\infty}^{\infty} e^{-u^2} \, du=\sqrt{\pi}$.
A: $x = \frac{1}{y}$ :
$$\int_0^\infty e^{-(x-\frac{1}{x})^2}dx = \int_0^\infty e^{-(\frac{1}{y}-y)^2}\frac{dy}{y^2}$$
$t =x-\frac{1}{x}, dt =(1+\frac{1}{x^2})dx $ :
$$\int_0^\infty e^{-(x-\frac{1}{x})^2}dx = \frac12\int_0^\infty e^{-(x-\frac{1}{x})^2}(1+\frac{1}{x^2})dx = \frac12\int_{-\infty}^\infty e^{-t^2}dt = \frac{\sqrt{\pi}}{2}$$
