How to show that the Laplace transform of $\exp(-t^2)$ is $\frac{\sqrt{\pi}}{2}\exp(\frac{s^2}{4})\rm erfc(\frac{s}{2})$

Question

I obtained the answer from Maple. But still I want to know how it is derived.

Answer 1

$\mathcal{L}\left\{e^{-t^2}\right\}$

$=\int_0^\infty e^{-t^2-st}~dt$

$=\int_0^\infty e^{-(t^2+st)}~dt$

$=\int_0^\infty e^{-\left(t^2+st+\frac{s^2}{4}-\frac{s^2}{4}\right)}~dt$

$=e^\frac{s^2}{4}\int_0^\infty e^{-\left(t+\frac{s}{2}\right)^2}~dt$

$=e^\frac{s^2}{4}\int_\frac{s}{2}^\infty e^{-t^2}~dt$

$=\dfrac{\sqrt\pi}{2}e^\frac{s^2}{4}~\text{erfc}\left(\dfrac{s}{2}\right)$