using convolution to find Laplace transform

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

$e^{\alpha t}\curvearrowleft {\displaystyle \frac{1}{s-\alpha }},{\displaystyle \frac{1}{\sqrt{t}}}\curvearrowleft \sqrt{{\displaystyle \frac{\pi }{s}}}\mathit{\hspace{1em}\hspace{1em}}(s>\alpha )$ (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

$$\frac{1}{(s-a^2)\sqrt{s}}\curvearrowright \frac{e^{a^{2}t}}{\sqrt{pi}}{\int }_0^{a\sqrt{t}}{e}^{-x^2}\cdot \frac{a}{x}\cdot \frac{2x}{a^2}𝑑x=\frac{e^{a^{2}t}}{a}\cdot \frac{2}{\sqrt{\pi }}{\int }_0^{a\sqrt{t}}{e}^{-x^2}𝑑x=\frac{e^{a^{2}t}}{a}\mathrm{erf}a\sqrt{t}.$$

Thus we may write the formula

$ℒ\{e^{a^{2}t}\mathrm{erf}a\sqrt{t}\}={\displaystyle \frac{a}{(s-a^2)\sqrt{s}}}\mathit{\hspace{1em}\hspace{1em}}(s>a^2).$ (2)

Moreover, we obtain

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

whence we have the other formula

$ℒ\{e^{a^{2}t}\mathrm{erfc}a\sqrt{t}\}={\displaystyle \frac{1}{(a+\sqrt{s})\sqrt{s}}}.$ (3)

0.1 An improper integral

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

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{a^2+x^2}𝑑x.$$

We have

$$e^{-t{x}^2}\curvearrowleft \frac{1}{s+x^2}$$

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

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{e^{-t{x}^2}}{a^2+x^2}}𝑑x$ $\mathrm{\hspace{0.33em}}\curvearrowleft {\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{dx}{(a^2+x^2)(s+x^2)}}={\displaystyle \frac{1}{s-a^2}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{a^2+x^2}}-{\displaystyle \frac{1}{s+x^2}}\right)𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{s-a^2}}\underset{x=0}{\overset{\mathrm{\infty }}{/}}\left({\displaystyle \frac{1}{a}}\arctan {\displaystyle \frac{x}{a}}-{\displaystyle \frac{1}{\sqrt{s}}}\arctan {\displaystyle \frac{x}{\sqrt{s}}}\right)$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{s-a^2}}\cdot {\displaystyle \frac{\pi }{2}}\left({\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{\sqrt{s}}}\right)={\displaystyle \frac{\pi }{2a}}\cdot {\displaystyle \frac{1}{(a+\sqrt{s})\sqrt{s}}}$

$\mathrm{\hspace{0.33em}}\curvearrowright {\displaystyle \frac{\pi }{2a}}{e}^{a^{2}t}\mathrm{erfc}a\sqrt{t}$

Consequently,

$${\int }_0^{\mathrm{\infty }}\frac{e^{-t{x}^2}}{a^2+x^2}𝑑x=\frac{\pi }{2a}{e}^{a^{2}t}\mathrm{erfc}a\sqrt{t},$$

and especially

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{a^2+x^2}𝑑x=\frac{\pi }{2a}{e}^{a^2}\mathrm{erfc}a.$$

Title	using convolution to find Laplace transform
Canonical name	UsingConvolutionToFindLaplaceTransform
Date of creation	2013-03-22 18:44:05
Last modified on	2013-03-22 18:44:05
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	12
Author	pahio (2872)
Entry type	Example
Classification	msc 26A42
Classification	msc 44A10
Related topic	ErrorFunction
Related topic	SubstitutionNotation
Related topic	IntegrationOfLaplaceTransformWithRespectToParameter