# using convolution to find Laplace transform

 $\displaystyle e^{\alpha t}\;\curvearrowleft\;\frac{1}{s\!-\!\alpha},\quad\frac% {1}{\sqrt{t}}\;\curvearrowleft\;\sqrt{\frac{\pi}{s}}\qquad(s>\alpha)$ (1)

where the curved 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

 $\frac{1}{(s\!-\!a^{2})\sqrt{s}}\;\;\curvearrowright\;\;e^{a^{2}t}*\frac{1}{% \sqrt{\pi t}}\;=\;\int_{0}^{t}\!e^{a^{2}(t-u)}\frac{1}{\sqrt{\pi u}}\,du.$

The substitution (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)  $a^{2}u=x^{2}$  then gives

 $\frac{1}{(s\!-\!a^{2})\sqrt{s}}\;\curvearrowright\;\frac{e^{a^{2}t}}{\sqrt{pi}% }\int_{0}^{a\sqrt{t}}\!e^{-x^{2}}\!\cdot\!\frac{a}{x}\!\cdot\!\frac{2x}{a^{2}}% \,dx\;=\;\frac{e^{a^{2}t}}{a}\!\cdot\!\frac{2}{\sqrt{\pi}}\int_{0}^{a\sqrt{t}}% \!e^{-x^{2}}\,dx\;=\;\frac{e^{a^{2}t}}{a}\,{\rm erf}\,a\sqrt{t}.$

Thus we may write the formula

 $\displaystyle\mathcal{L}\{e^{a^{2}t}\,{\rm erf}\,a\sqrt{t}\}\;=\;\frac{a}{(s\!% -\!a^{2})\sqrt{s}}\qquad(s>a^{2}).$ (2)

Moreover, we obtain

 $\frac{1}{(\sqrt{s}\!+\!a)\sqrt{s}}\;=\;\frac{\sqrt{s}\!-\!a}{(s\!-\!a^{2})% \sqrt{s}}\;=\,\frac{1}{s-a^{2}}-\frac{a}{(s-a^{2})\sqrt{s}}\;\curvearrowright% \;e^{a^{2}t}-e^{a^{2}t}\,{\rm erf}\,a\sqrt{t}\;=\;e^{a^{2}t}(1-{\rm erf}\,a% \sqrt{t}),$

whence we have the other formula

 $\displaystyle\mathcal{L}\{e^{a^{2}t}\,{\rm erfc}\,a\sqrt{t}\}\;=\;\frac{1}{(a% \!+\!\sqrt{s})\sqrt{s}}.$ (3)

## 0.1 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 table of Laplace transforms (http://planetmath.org/TableOfLaplaceTransforms)).  Dividing this by $a^{2}\!+\!x^{2}$ and integrating from 0 to $\infty$, we can continue as follows:

 $\displaystyle\int_{0}^{\infty}\frac{e^{-tx^{2}}}{a^{2}\!+\!x^{2}}\,dx$ $\displaystyle\;\curvearrowleft\;\int_{0}^{\infty}\frac{dx}{(a^{2}\!+\!x^{2})(s% \!+\!x^{2})}\;=\;\frac{1}{s\!-\!a^{2}}\int_{0}^{\infty}\left(\frac{1}{a^{2}\!+% \!x^{2}}-\frac{1}{s\!+\!x^{2}}\right)dx$ $\displaystyle\;=\;\frac{1}{s\!-\!a^{2}}\operatornamewithlimits{\Big{/}}_{\!\!% \!x=0}^{\,\quad\infty}\left(\frac{1}{a}\arctan\frac{x}{a}-\frac{1}{\sqrt{s}}% \arctan\frac{x}{\sqrt{s}}\right)$ $\displaystyle\;=\;\frac{1}{s\!-\!a^{2}}\!\cdot\!\frac{\pi}{2}\left(\frac{1}{a}% -\frac{1}{\sqrt{s}}\right)\;=\;\frac{\pi}{2a}\!\cdot\!\frac{1}{(a\!+\!\sqrt{s}% )\sqrt{s}}$ $\displaystyle\;\curvearrowright\;\frac{\pi}{2a}e^{a^{2}t}\,{\rm erfc}\,a\sqrt{t}$

Consequently,

 $\int_{0}^{\infty}\frac{e^{-tx^{2}}}{a^{2}\!+\!x^{2}}\,dx\;=\;\frac{\pi}{2a}e^{% a^{2}t}\,{\rm erfc}\,a\sqrt{t},$

and especially

 $\int_{0}^{\infty}\frac{e^{-x^{2}}}{a^{2}\!+\!x^{2}}\,dx\;=\;\frac{\pi}{2a}e^{a% ^{2}}\,{\rm erfc}\,a.$
Title using convolution  to find Laplace transform UsingConvolutionToFindLaplaceTransform 2013-03-22 18:44:05 2013-03-22 18:44:05 pahio (2872) pahio (2872) 12 pahio (2872) Example msc 26A42 msc 44A10 ErrorFunction SubstitutionNotation IntegrationOfLaplaceTransformWithRespectToParameter