what is the closed form for the inverse Laplace transform of Gaussian function $\mathcal{L^{-1}}\left\{e^{-x^2}\right\}$?

Question

it's easy to get the laplace transform of Gaussian function as the following:

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

$=\int_0^\infty e^{-x^2-sx}~dx$

$=\int_0^\infty e^{-(x^2+sx)}~dx$

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

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

Now my question here what is the inverse Laplace transform of it since WA assumed that there is no result in terms of standard mathematical functions?

Answer 1

$\sqrt{\pi /2} e^{s^2/2}$ is the bilateral Laplace transform of $e^{-t^2/2}$ which is not the same as $e^{-t^2/2} 1_{t > 0}$

$e^{-s^2}$ grows way too fast on vertical lines to be the Laplace transform of a tempered distribution.