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# Laplace transform of a Gaussian function

We evaluate the Laplace transform
^{1}^{1}cf. *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.

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