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[parent] using convolution to find Laplace transform (Example)

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

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

An improper integral

One can utilise the formula (3) for evaluating the improper integral $$\int_0^\infty\frac{e^{-x^2}}{a^2\!+\!x^2}\,dx.$$ 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:

$\displaystyle \int_0^\infty\frac{e^{-tx^2}}{a^2\!+\!x^2}\,dx$ $\displaystyle \;\curvearrowleft\; \int_0^\infty\frac{dx}{(a^2\!+\!x^2)(s\!+\!x^... ...s\!-\!a^2}\int_0^\infty\left(\frac{1}{a^2\!+\!x^2}-\frac{1}{s\!+\!x^2}\right)dx$    
  $\displaystyle \;=\; \frac{1}{s\!-\!a^2}\operatornamewithlimits{\Big/}_{\!\!\!x=... ...frac{1}{a}\arctan\frac{x}{a}-\frac{1}{\sqrt{s}}\arctan\frac{x}{\sqrt{s}}\right)$    
  $\displaystyle \;=\; \frac{1}{s\!-\!a^2}\!\cdot\!\frac{\pi}{2}\left(\frac{1}{a}-... ...sqrt{s}}\right) \;=\; \frac{\pi}{2a}\!\cdot\!\frac{1}{(a\!+\!\sqrt{s})\sqrt{s}}$    
  $\displaystyle \;\curvearrowright\; \frac{\pi}{2a}e^{a^2t}\,{\rm erfc}\,a\sqrt{t}$    

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




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See Also: error function, substitution notation, integration of Laplace transform with respect to parameter


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Cross-references: improper integral, convolution property of Laplace transform, functions, table of Laplace transforms
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This is version 9 of using convolution to find Laplace transform, born on 2009-01-14, modified 2009-05-05.
Object id is 11504, canonical name is UsingConvolutionToFindLaplaceTransform.
Accessed 896 times total.

Classification:
AMS MSC44A10 (Integral transforms, operational calculus :: Laplace transform)
 26A42 (Real functions :: Functions of one variable :: Integrals of Riemann, Stieltjes and Lebesgue type)

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