I am trying to figure out how to perform the following computation. The objective is to compute the laplace transform of the smirnov density. The lecture notes I've seen online state that
$\displaystyle \int_0^{\infty} e^{-st} \frac{x}{\sqrt{2\pi}} t^{-3/2}e^{-x^2/2t}\,dt = e^{x \sqrt{2s}}$ where $x,s >0$.
I'm a bit stuck on the computation and was wondering if anyone can provide me with some hints as to how to proceed. Since $x,s$ are constants above then by appropriately choosing their values, and taking out the constants from the integrand, we obtain
$\displaystyle \int_0^{\infty} e^{-t - 1/t} t^{-3/2} \,dt$ which is much easier to work with.
After plugging this into wolfram alpha, I got something related to the error function so in anticipation of that, I did:
let $u = t^{-1/2}$ then taking the constants out (sorry it's just effort to type them in), the above integral reduces to:
$\displaystyle \int_0^{\infty} e^{-u^2 - 1/u^2} \,du$.
By multiplying the integrand by $e^{2-2}$, I can rewrite it as (and taking out the constants again):
$\displaystyle \int_0^{\infty} e^{-(u+1/u)^2} \,du$.
I am currently stuck at this point and am tempted to try $z = u + 1/u$ but it results in a more complicated expression.
This is a calculation at the last step of a hw problem so I just need some hints/suggestions if I'm on the right track.
Thanks a lot.
Hint: $$\int_{-\infty}^\infty f(x)\,dx=\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,dx$$
Edit: See this page for proofs of the identity.
Of course, you need to establish that your function is even in order to take advantage of this identity.