Laplace transform of a Gaussian function

We evaluate the Laplace transform

$ℒ\{e^{-t^2}\}={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-st}{e}^{-t^2}𝑑t=F(s).$ (1)

In fact,

$ℒ\{e^{-t^2}\}={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-(t^2+2\frac{s}{2}t+\frac{s^2}{4}-\frac{s^2}{4})}𝑑t=e^{\frac{s^2}{4}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-{(t+\frac{s}{2})}^2}𝑑t.$

By making the change of variable $t+\frac{s}{2}=u$, we have (by the second equality in (1), the variable on operator’s argument is immaterial)

$ℒ\{e^{-t^2}\}=e^{\frac{s^2}{4}}{\displaystyle {\int }_{\frac{s}{2}}^{\mathrm{\infty }}}{e}^{-u^2}𝑑u.$

That is,

$ℒ\{e^{-t^2}\}=F(s)={\displaystyle \frac{\sqrt{\pi }}{2}}{e}^{\frac{s^2}{4}}\mathrm{erfc}\left({\displaystyle \frac{s}{2}}\right),$

where $\mathrm{erfc}(\cdot )$ is the complementary error function. Its path of integration is subject to the restriction $\arg (u)\to \theta$, with $|\theta |\le \pi /4$ as $u\to \mathrm{\infty }$ along the path, with equality only if $\mathrm{\Re }(u^2)$ remains bounded to the left.

Title	Laplace transform of a Gaussian function
Canonical name	LaplaceTransformOfAGaussianFunction
Date of creation	2013-03-22 16:03:21
Last modified on	2013-03-22 16:03:21
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	5
Author	perucho (2192)
Entry type	Application
Classification	msc 42-01