Laplace Transform of $\frac{a}{2\sqrt{\pi}}t^{-3/2}\exp(-a^2/(4t))$

Question

I'm trying to prove that the Laplace Transform of

$$\frac{a} {2\sqrt{\pi}}t^{-3/2}\exp(-a^2/(4t)) $$

is

$$\exp(-a\sqrt{s}); $$

from the definition of Laplace Transform we should compute the following integral: $\int_{0}^{\infty}\exp(-st-\frac{1}{t})t^{-3/2}dt$, but atm I am stuck with it...

Can you provide some hints or reference links to the worked out computation? Thanks.

Answer 1

Here is an approach. The integral in consideration is

$$ \int_{0}^{\infty}t^{-3/2} e^{-\frac{a^2}{4t}}e^{-st}dt. $$

Taking the Mellin transform w.r.t. $a$ (see note 1), we need to change the order of integration, gives

$$ \int_{0}^{\infty}t^{-3/2}e^{-st} \int_{0}^{\infty} a^{w-1}e^{-\frac{a^2}{4t}} da\,dt = \frac{4^{-w/2}}{2}\,\Gamma\left( -\frac{w}{2} \right) \int_{0}^{\infty}t^{-w/2-3/2}e^{-st} dt$$

$$ = 2\,\sqrt {\pi }\,{s}^{w/2+1/2}\Gamma( -w-1 ) =2\sqrt {\pi }\,{s}^{u/2}\Gamma\left( -u \right),\quad ( w=u-1 ). $$

The last integral is easy to evaluate using the gamma function (see note 3, with using the change of variables $st=z$). Now to finish the problem you need to find the inverse Mellin transform w.r.t $u$ (see note 2) which gives the desired result

$$ \frac{a}{2\sqrt{\pi}} \int_{0}^{\infty}t^{-3/2} e^{-\frac{a^2}{4t}}e^{-st}dt = e^{-a\sqrt{s}}.$$

Notes:

1) The Mellin transfor of $e^{-\alpha x^2}$ is

$$ \frac{\alpha^{w/2}}{2}\Gamma\left(-\frac{w}{2}\right) $$

2) To find the inverse Mellin transform just notice that the Mellin transform of $e^{-cx}$ is given by (see tables of Mellin transform)

$$ c^{w}\Gamma(-w). $$

3) The gamma function is defined as

$$ \Gamma( z ) = \int_{0}^{\infty} t^{z-1} e^{-t}\,dt. $$