PlanetPhysics/Using Convolution to Find Laplace Transforms

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

where the curved arrows point from the Laplace-transformed functions to the original functions.\, Setting\, \, and dividing by in (1), the convolution property of Laplace transform yields The substitution \,\, then gives

Thus we may write the formula

Moreover, we obtain whence we have the other formula

An improper integral

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One can utilise the formula (3) for evaluating the improper integral   We have   (see the table of Laplace transforms).\, Dividing this by   and integrating from 0 to  , 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,   and especially