PlanetPhysics/Using Convolution to Find Laplace Transforms

We start from the relations (see the [[../2DLT/|table of Laplace transforms]])

$\begin{array}{c}{e}^{\alpha t}\curvearrowleft \frac{1}{s-\alpha },\frac{1}{\sqrt{t}}\curvearrowleft \sqrt{\frac{\pi }{s}}(s>\alpha )\end{array}$

where the curved arrows point from the Laplace-transformed [[../Bijective/|functions]] to the original functions.\, Setting\, $\alpha =a^2$\, and dividing by $\sqrt{\pi }$ in (1), the [[../688/|convolution property]] of [[../2DLT/|Laplace transform]] yields $\frac{1}{(s-a^2)\sqrt{s}}\curvearrowright e^{a^{2}t}*\frac{1}{\sqrt{\pi t}}=\underset{0}{\overset{t}{\int }}{e}^{a^2(t-u)}\frac{1}{\sqrt{\pi u}}du.$ The substitution \,$a^{2}u=x^2$\, then gives $\frac{1}{(s-a^2)\sqrt{s}}\curvearrowright \frac{e^{a^{2}t}}{\sqrt{pi}}\underset{0}{\overset{a\sqrt{t}}{\int }}{e}^{-x^2}\cdot \frac{a}{x}\cdot \frac{2x}{a^2}dx=\frac{e^{a^{2}t}}{a}\cdot \frac{2}{\sqrt{\pi }}\underset{0}{\overset{a\sqrt{t}}{\int }}{e}^{-x^2}dx=\frac{e^{a^{2}t}}{a}\mathrm{e}\mathrm{r}\mathrm{f}a\sqrt{t}.$

Thus we may write the [[../Formula/|formula]]

$\begin{array}{c}ℒ\{e^{a^{2}t}\mathrm{e}\mathrm{r}\mathrm{f}a\sqrt{t}\}=\frac{a}{(s-a^2)\sqrt{s}}(s>a^2).\end{array}$

Moreover, we obtain $\frac{1}{(\sqrt{s}+a)\sqrt{s}}=\frac{\sqrt{s}-a}{(s-a^2)\sqrt{s}}=\frac{1}{s-a^2}-\frac{a}{(s-a^2)\sqrt{s}}\curvearrowright e^{a^{2}t}-e^{a^{2}t}\mathrm{e}\mathrm{r}\mathrm{f}a\sqrt{t}=e^{a^{2}t}(1-\mathrm{e}\mathrm{r}\mathrm{f}a\sqrt{t}),$ whence we have the other formula

$\begin{array}{c}ℒ\{e^{a^{2}t}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{c}a\sqrt{t}\}=\frac{1}{(a+\sqrt{s})\sqrt{s}}.\end{array}$

An improper integral

One can utilise the formula (3) for evaluating the improper integral $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}}{a^2+x^2}dx.$ We have $e^{-t{x}^2}\curvearrowleft \frac{1}{s+x^2}$ (see the [[../TableOfLaplaceTransforms/|table of Laplace transforms]]).\, Dividing this by $a^2+x^2$ and integrating from 0 to $\mathrm{\infty }$, 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, $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-t{x}^2}}{a^2+x^2}dx=\frac{\pi }{2a}{e}^{a^{2}t}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{c}a\sqrt{t},$ and especially $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}}{a^2+x^2}dx=\frac{\pi }{2a}{e}^{a^2}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{c}a.$