Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$. Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$.
I'm not really sure of how I should proceed to show this, and it's pretty un-intuitive as well.
I've managed to manipulate the RHS into 
$$\frac{e^{4\beta}}{\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{u}} e^{-(\beta+2u)^2/4u} du$$
which now vaguely resembles a Gaussian. However, substituting $v = \beta + 2u$ in in order to to get something like $e^{-x^2}$ doesn't particularly help. I got
$$
\frac{e^{4\beta}}{\sqrt{2\pi}} \int_0^\infty \frac{1}{\sqrt{v - \beta}} e^{-v^2/2(v-\beta)} dv.
$$
I'm really unsure of how I should proceed; any help would be appreciated.

Hint. By the change of variable $u=x^2$ we obtain
$$
\int_0^{\infty}\frac{e^{-(u+\frac{b^2}{u})}}{\sqrt u}dt=2\int_0^{\infty}e^{ -x^2-b^2/x^2}dx=\int_{-\infty}^{\infty}e^{ -x^2-b^2/x^2}dx
$$
one may then recall that, for any integrable function $f$, we have (see here)

$$
\int_{-\infty}^{+\infty}f\left(x-\frac{b}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad b>0. \tag1
$$

Applying it to $f(x)=e^{-x^2}$, gives
$$
\int_{-\infty}^{+\infty}e^{-(x-b/x)^2}\mathrm{d}x=\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x=\sqrt{\pi}, \quad b>0. \tag2
$$
Thus
$$
\int_{-\infty}^{+\infty}e^{-x^2-b^2/x^{2}}\mathrm{d}x=\sqrt{\pi}\:e^{-2b}\tag3
$$ from which it is clear how to deduce the announced result.
