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.
 A: 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.
A: Consider the integral $$k(\beta,s)=\int_0^\infty e^{-s u}\frac{e^{-\beta^2/4u}}{\sqrt{4\pi u}} \,dt,$$ the Laplace transform of $K(\beta,u)=\frac{1}{\sqrt{4\pi u}}e^{-\beta^2/4u}$. The significance of $K(\beta,u)$ is that it's the 1D heat kernel with position $\beta$ and time $u$. That is, it satisfies the heat equation $$\frac{\partial K}{\partial u}=\frac{\partial^2 K}{\partial \beta^2}$$ for all $u>0$ and converges to a Dirac delta $\delta(\beta)$ as $u\to 0$. (One may check that, for all $u>0$, $K(\beta,u)$ defines a Gaussian PDF with zero mean and variance $\sigma^2=u$.)
To make use of this fact, we take the Laplace transform of both sides of the heat equation:
\begin{align} 
\int_0^\infty e^{-s u} \frac{\partial^2}{\partial\beta^2}K(\beta,u)\,du
&=\frac{\partial^2}{\partial\beta^2}\int_0^\infty e^{-s u} K(\beta,u)\,du=\frac{\partial^2 k}{\partial\beta^2}\\
&=\int_0^\infty e^{-s u} \frac{\partial}{\partial u}K(\beta,u)\,du\\
&=\left[e^{-s u}K(\beta,u)\right]^{\infty}_{u=0}+\int_0^\infty se^{-s u} K(\beta,u)\,du\\
&=-\delta(\beta)+s k(\beta,s)
\end{align}
where in the second-to-last line we have integrated by parts and subsequently used the boundary condition $K(\beta,u)\to \delta(\beta)$ as $u\to 0$. Regarding $s$ now as a parameter, we have the ODE $k''(\beta)=s\, k(\beta)-\delta(\beta)$. This may be solved to give $k(\beta,s)=\frac{1}{2}e^{-|\beta|\sqrt{s}}$. (Briefly: For $\beta\neq 0$, the equation is just $k''(\beta)=sk(\beta)$ with $k(\beta)$ vanishing at infinity; to account for the delta function, integrate both sides of the ODE from $-\epsilon$ to $\epsilon$ and take $\epsilon\to 0^+$ to obtain $k'(0^+)-k'(0^-)=1$.) Taking $s\to 1$ and assuming $\beta>0$, we obtain the announced result $2k(\beta,1)=e^{-\beta}$.
