Show that $\int_0^\infty e^{-t^2-x^2/t}t^{-1/2}dt = x\int_0^\infty e^{-t^2-x^2/t}t^{-3/2}dt$ For $x>0$, using a substitution, show that 
$\int_0^\infty e^{-t^2-x^2/t}t^{-1/2}dt = x\int_0^\infty e^{-t^2-x^2/t}t^{-3/2}dt$.
My attempt: For the LHS, if we use $u=\sqrt{t}$, $du=\frac{1}{2}t^{-\frac{1}{2}}$, and the integral turns into
$\frac{2}{e}\int_0^\infty e^{(\frac{x}{u})^2}du$.
For the RHS, using $u=t^{-1/2}$, $du=\frac{-1}{2}t^{-3/2}$, we get
$\frac{x}{e}\int_0^\infty e^{-(xu)^2}du$.
I now want to show these are equal, but I'm not sure how.
Any help appreciated!

More of an observation. 
$$
\frac{d}{dx}\int_0^\infty\mathrm{e}^{-t^2-\frac{x^2}{t}}t^{-1/2}dt = -2x\int_0^\infty \mathrm{e}^{-t^2-\frac{x^2}{t}}t^{-3/2}dt
$$
so your equality is essentially (assuming that your equation is true)
$$
\frac{dI}{dx} = -2I  
$$
which means
$$
\int_0^\infty\mathrm{e}^{-t^2-\frac{x^2}{t}}t^{-1/2}dt = \frac{k}{x^2}
$$
How does this match with what you see?
