UsingConvolutionToFindLaplaceTransforms 2009-05-04 18:05:27 2009-05-04 18:05:27 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sijoitus}[2]% {\operatornamewithlimits{\Big/}_{\!\!\!#1}^{\,#2}} We start from the \PMlinkescapetext{relations} (see the table of Laplace transforms) \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 \PMlinkescapetext{arrows point} from the Laplace-transformed functions to the original functions.\, Setting\, $\alpha = a^2$\, and dividing by $\sqrt{\pi}$ in (1), the convolution property of Laplace transform 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 \PMlinkname{substitution}{ChangeOfVariableInDefiniteIntegral} \,$a^2u = x^2$\, 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 formula \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 $$\int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}.$$ We have $$e^{-tx^2} \;\curvearrowleft\; \frac{1}{s\!+\!x^2}$$ (see the \PMlinkname{table of Laplace transforms}{TableOfLaplaceTransforms}).\, 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 $$\int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}\,dx \;=\; \frac{\pi}{2a}e^{a^2}\,{\rm erfc}\,a.$$