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using convolution to find Laplace transforms (Definition)

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

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

$\displaystyle \frac{1}{(s\!-\!a^2)\sqrt{s}} \;\;\curvearrowright\;\; e^{a^2t}*\frac{1}{\sqrt{\pi t}} \;=\; \int_0^t\!e^{a^2(t-u)}\frac{1}{\sqrt{\pi u}}\,du.$
The substitution   $a^2u = x^2$  then gives

$\displaystyle \frac{1}{(s\!-\!a^2)\sqrt{s}} \;\curvearrowright\; \frac{e^{a^2t}... ...int_0^{a\sqrt{t}}\!e^{-x^2}\,dx \;=\; \frac{e^{a^2t}}{a}\,{\rm erf}\,a\sqrt{t}.$
Thus we may write the formula
$\displaystyle \mathcal{L}\{e^{a^2t}\,{\rm erf}\,a\sqrt{t}\} \;=\; \frac{a}{(s\!-\!a^2)\sqrt{s}} \qquad (s > a^2).$ (2)

Moreover, we obtain

$\displaystyle \frac{1}{(\sqrt{s}\!+\!a)\sqrt{s}} \;=\; \frac{\sqrt{s}\!-\!a}{(s... ...e^{a^2t}-e^{a^2t}\,{\rm erf}\,a\sqrt{t} \;=\; e^{a^2t}(1-{\rm erf}\,a\sqrt{t}),$
whence we have the other formula
$\displaystyle \mathcal{L}\{e^{a^2t}\,{\rm erfc}\,a\sqrt{t}\} \;=\; \frac{1}{(a\!+\!\sqrt{s})\sqrt{s}}.$ (3)

An improper integral

One can utilise the formula (3) for evaluating the improper integral

$\displaystyle \int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}\,dx.$
We have

$\displaystyle e^{-tx^2} \;\curvearrowleft\; \frac{1}{s\!+\!x^2}$
(see the table of Laplace transforms).  Dividing this by $a^2\!+\!x^2$ and integrating from 0 to $\infty$, we can continue as follows:
$\displaystyle \int_0^\infty\frac{e^{-tx^2}}{a^2\!+\!x^2}\,dx$ $\displaystyle \;\curvearrowleft\; \int_0^\infty\frac{dx}{(a^2\!+\!x^2)(s\!+\!x^... ...s\!-\!a^2}\int_0^\infty\left(\frac{1}{a^2\!+\!x^2}-\frac{1}{s\!+\!x^2}\right)dx$    
  $\displaystyle \;=\; \frac{1}{s\!-\!a^2}\operatornamewithlimits{\Big/}_{\!\!\!x=... ...frac{1}{a}\arctan\frac{x}{a}-\frac{1}{\sqrt{s}}\arctan\frac{x}{\sqrt{s}}\right)$    
  $\displaystyle \;=\; \frac{1}{s\!-\!a^2}\!\cdot\!\frac{\pi}{2}\left(\frac{1}{a}-... ...sqrt{s}}\right) \;=\; \frac{\pi}{2a}\!\cdot\!\frac{1}{(a\!+\!\sqrt{s})\sqrt{s}}$    
  $\displaystyle \;\curvearrowright\; \frac{\pi}{2a}e^{a^2t}\,{\rm erfc}\,a\sqrt{t}$    
Consequently,

$\displaystyle \int_0^\infty\frac{e^{-tx^2}}{a^2\!+\!x^2}\,dx \;=\; \frac{\pi}{2a}e^{a^2t}\,{\rm erfc}\,a\sqrt{t},$
and especially

$\displaystyle \int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}\,dx \;=\; \frac{\pi}{2a}e^{a^2}\,{\rm erfc}\,a.$



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See Also: table of Laplace transforms


Cross-references: formula, Laplace transform, functions, table of Laplace transforms

This is version 3 of using convolution to find Laplace transforms, born on 2009-05-04, modified 2009-05-05.
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Physics Classification: 02.30.Uu (Integral transforms)
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