using convolution to find Laplace transforms

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

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

Anyone with an account can edit this entry. Please help improve it!

"using convolution to find Laplace transforms" is owned by pahio.
(view preamble)