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

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