Laplace Transform of $\frac{e^{\frac{-a^2}{4t}}}{\sqrt{\pi t}}$.

Question

I would like the Laplace Transform of $$\frac{e^{\frac{-a^2}{4t}}}{\sqrt{\pi t}}.$$

Using the definition of a Laplace transform I need to evaluate the integral $$\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}\exp\left(\frac{-a^2}{4u^2}-su^2\right)du$$

This is where I'm stuck. I've been given that $$\int_{0}^{\infty}\exp\left(\frac{-c^2}{x^2}-c^2x^2\right)dx = \frac{\sqrt{\pi}}{2|c|}e^{-2c^2}$$but I don't know what value of c to choose. Any help?

Answer 1

Assume $s>0$ and $a\ne0.$ In order to have $$\frac{-a^2}{4u^2}=\frac{-c^2}{x^2}\quad\text{and}\quad su^2=c^2x^2,$$ choose $$c=\sqrt{\frac{|a|\sqrt s}2}\quad\text{and}\quad x=u\frac{\sqrt s}c.$$ Using the identity you have been given, i.e. $$\int_{0}^{\infty}\exp\left(\frac{-c^2}{x^2}-c^2x^2\right)dx = \frac{\sqrt{\pi}}{2|c|}e^{-2c^2},$$ you get: $$\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}\exp\left(\frac{-a^2}{4u^2}-su^2\right)du=\frac1{\sqrt s}e^{-|a|\sqrt s}.$$ By uniqueness of the analytic continuation, $\mathcal{L}_t\left[\frac{\exp \left(-\frac{a^2}{4 t}\right)}{\sqrt{\pi t}}\right](s)=\frac1{\sqrt s}e^{-|a|\sqrt s}$ holds not only for every real number $s>0$ but for every complex number $s$ such that $\Re(s)>0.$

Answer 2

First let $t=su^2$ which gives that

$$ I = \int_0^{\infty} \exp\left( -\frac{a^2}{4u^2}-su^2\right) du = \frac{1}{2\sqrt{s}} \int_0^{\infty} \exp\left(\frac{sa^2}{4t}-t \right) \frac{dt}{t^{1/2}} $$

This is a form of a Bessel function which gives

$$ \int_0^{\infty} \exp\left(-\frac{sa^2}{4t}-t \right) \frac{dt}{t^{1/2}} = \sqrt{2a\sqrt{s}}K_{1/2}(a\sqrt{s}) $$

such that we have

$$ I = \frac{\sqrt{a\sqrt{s}}}{\sqrt{2s}}K_{1/2}(a\sqrt{s}) $$

Since

$$K_{1/2}(a\sqrt{s}) = \frac{\sqrt{\frac{\pi}{2}}e^{-a\sqrt{s}}}{\sqrt{a\sqrt{s}}} $$

we may simplify to obtain

$$ I = \sqrt{\frac{\pi}{2}} \frac{e^{-a\sqrt{s}}}{\sqrt{2s}} $$

which in the end gives that the Laplace transform is

$$ \frac{2}{\sqrt{\pi}} \int_0^{\infty} \exp\left( -\frac{a^2}{4u^2}-su^2\right) du = \frac{e^{-a\sqrt{s}}}{\sqrt{s}} $$

as mentioned in the hint. I've assumed $a>0$ which is possible to extend to negative values as well, which then gives the absolute value.