# Laplace transform of a Gaussian function

We evaluate the Laplace transform 11cf. Gaussian function, wikipedia.org

 $\displaystyle\mathcal{L}\{e^{-t^{2}}\}=\int_{0}^{\infty}e^{-st}e^{-t^{2}}\,dt=% F(s).$ (1)

In fact,

 $\displaystyle\mathcal{L}\{e^{-t^{2}}\}=\int_{0}^{\infty}e^{-(t^{2}+2\frac{s}{2% }t+\frac{s^{2}}{4}-\frac{s^{2}}{4})}\,dt=e^{\frac{s^{2}}{4}}\!\!\int_{0}^{% \infty}e^{-(t+\frac{s}{2})^{2}}\,dt.$

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)

 $\displaystyle\mathcal{L}\{e^{-t^{2}}\}=e^{\frac{s^{2}}{4}}\!\!\int_{\frac{s}{2% }}^{\infty}e^{-u^{2}}\,du.$

That is,

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

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

Title Laplace transform of a Gaussian function LaplaceTransformOfAGaussianFunction 2013-03-22 16:03:21 2013-03-22 16:03:21 perucho (2192) perucho (2192) 5 perucho (2192) Application msc 42-01