PlanetPhysics/Using Convolution to Find Laplace Transforms

We start from the relations (see the table of Laplace transforms)

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

where the curved 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 ${\frac {1}{(s\!-\!a^{2}){\sqrt {s}}}}\;\;\curvearrowright \;\;e^{a^{2}t}*{\frac {1}{\sqrt {\pi t}}}\;=\;\int _{0}^{t}\!e^{a^{2}(t-u)}{\frac {1}{\sqrt {\pi u}}}\,du.$ The substitution \,$a^{2}u=x^{2}$\, 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

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

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

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

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).\, Dividing this by $a^{2}\!+\!x^{2}$ and integrating from 0 to $\infty$, we can continue as follows:

Failed to parse (unknown function "\sijoitus"): {\displaystyle \begin{matrix} \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{matrix}}

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 $\int _{0}^{\infty }{\frac {e^{-x^{2}}}{a^{2}\!+\!x^{2}}}\,dx\;=\;{\frac {\pi }{2a}}e^{a^{2}}\,{\rm {erfc}}\,a.$