Talk:PlanetPhysics/Using Convolution to Find Laplace Transforms

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: using convolution to find Laplace transforms %%% Primary Category Code: 02.30.Uu %%% Filename: UsingConvolutionToFindLaplaceTransforms.tex %%% Version: 3 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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We start from the relations (see the \htmladdnormallink{table of Laplace transforms}{http://planetphysics.us/encyclopedia/2DLT.html}) \begin{align} e^{\alpha t} \;\curvearrowleft\; \frac{1}{s\!-\!\alpha}, \quad \frac{1}{\sqrt{t}} \;\curvearrowleft\; \sqrt{\frac{\pi}{s}} \qquad (s > \alpha) \end{align} where the curved arrows point from the Laplace-transformed \htmladdnormallink{functions}{http://planetphysics.us/encyclopedia/Bijective.html} to the original functions.\, Setting\, $\alpha = a^2$\, and dividing by $\sqrt{\pi}$ in (1), the \htmladdnormallink{convolution property}{http://planetphysics.us/encyclopedia/688.html} of \htmladdnormallink{Laplace transform}{http://planetphysics.us/encyclopedia/2DLT.html} yields $$\frac{1}{(s\!-\!a^2)\sqrt{s}} \;\;\curvearrowright\;\; e^{a^2t}*\frac{1}{\sqrt{\pi t}} \;=\; \int_0^t\!e^{a^2(t-u)}\frac{1}{\sqrt{\pi u}}\,du.$$ The substitution \,$a^2u = x^2$\, then gives $$\frac{1}{(s\!-\!a^2)\sqrt{s}} \;\curvearrowright\; \frac{e^{a^2t}}{\sqrt{pi}}\int_0^{a\sqrt{t}}\!e^{-x^2}\!\cdot\!\frac{a}{x}\!\cdot\!\frac{2x}{a^2}\,dx \;=\; \frac{e^{a^2t}}{a}\!\cdot\!\frac{2}{\sqrt{\pi}}\int_0^{a\sqrt{t}}\!e^{-x^2}\,dx \;=\; \frac{e^{a^2t}}{a}\,{\rm erf}\,a\sqrt{t}.$$ Thus we may write the \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html} \begin{align} \mathcal{L}\{e^{a^2t}\,{\rm erf}\,a\sqrt{t}\} \;=\; \frac{a}{(s\!-\!a^2)\sqrt{s}} \qquad (s > a^2). \end{align}

Moreover, we obtain $$\frac{1}{(\sqrt{s}\!+\!a)\sqrt{s}} \;=\; \frac{\sqrt{s}\!-\!a}{(s\!-\!a^2)\sqrt{s}} \;=\, \frac{1}{s-a^2}-\frac{a}{(s-a^2)\sqrt{s}} \;\curvearrowright\; e^{a^2t}-e^{a^2t}\,{\rm erf}\,a\sqrt{t} \;=\; e^{a^2t}(1-{\rm erf}\,a\sqrt{t}),$$ whence we have the other formula \begin{align} \mathcal{L}\{e^{a^2t}\,{\rm erfc}\,a\sqrt{t}\} \;=\; \frac{1}{(a\!+\!\sqrt{s})\sqrt{s}}. \end{align}

\subsection{An improper integral}

One can utilise the formula (3) for evaluating the improper integral $$\int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}\,dx.$$ We have $$e^{-tx^2} \;\curvearrowleft\; \frac{1}{s\!+\!x^2}$$ (see the \htmladdnormallink{table of Laplace transforms}{http://planetphysics.us/encyclopedia/TableOfLaplaceTransforms.html}).\, Dividing this by $a^2\!+\!x^2$ and integrating from 0 to $\infty$, we can continue as follows: \begin{align*} \int_0^\infty\frac{e^{-tx^2}}{a^2\!+\!x^2}\,dx & \;\curvearrowleft\; \int_0^\infty\frac{dx}{(a^2\!+\!x^2)(s\!+\!x^2)} \;=\; \frac{1}{s\!-\!a^2}\int_0^\infty\left(\frac{1}{a^2\!+\!x^2}-\frac{1}{s\!+\!x^2}\right)dx\\ & \;=\; \frac{1}{s\!-\!a^2}\sijoitus{x=0}{\quad\infty}\left(\frac{1}{a}\arctan\frac{x}{a}-\frac{1}{\sqrt{s}}\arctan\frac{x}{\sqrt{s}}\right)\\ & \;=\; \frac{1}{s\!-\!a^2}\!\cdot\!\frac{\pi}{2}\left(\frac{1}{a}-\frac{1}{\sqrt{s}}\right) \;=\; \frac{\pi}{2a}\!\cdot\!\frac{1}{(a\!+\!\sqrt{s})\sqrt{s}}\\ & \;\curvearrowright\; \frac{\pi}{2a}e^{a^2t}\,{\rm erfc}\,a\sqrt{t} \end{align*} Consequently, $$\int_0^\infty\frac{e^{-tx^2}}{a^2\!+\!x^2}\,dx \;=\; \frac{\pi}{2a}e^{a^2t}\,{\rm erfc}\,a\sqrt{t},$$ and especially $$\int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}\,dx \;=\; \frac{\pi}{2a}e^{a^2}\,{\rm erfc}\,a.$$

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