We start from the relations (see the table of Laplace transforms )
e
α
t
↶
1
s
−
α
,
1
t
↶
π
s
(
s
>
α
)
{\displaystyle {\begin{matrix}e^{\alpha t}\;\curvearrowleft \;{\frac {1}{s\!-\!\alpha }},\quad {\frac {1}{\sqrt {t}}}\;\curvearrowleft \;{\sqrt {\frac {\pi }{s}}}\qquad (s>\alpha )\end{matrix}}}
where the curved arrows point from the Laplace-transformed functions to the original functions.\, Setting\,
α
=
a
2
{\displaystyle \alpha =a^{2}}
π
{\displaystyle {\sqrt {\pi }}}
convolution property of Laplace transform yields
1
(
s
−
a
2
)
s
↷
e
a
2
t
∗
1
π
t
=
∫
0
t
e
a
2
(
t
−
u
)
1
π
u
d
u
.
{\displaystyle {\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.}
a
2
u
=
x
2
{\displaystyle a^{2}u=x^{2}}
1
(
s
−
a
2
)
s
↷
e
a
2
t
p
i
∫
0
a
t
e
−
x
2
⋅
a
x
⋅
2
x
a
2
d
x
=
e
a
2
t
a
⋅
2
π
∫
0
a
t
e
−
x
2
d
x
=
e
a
2
t
a
e
r
f
a
t
.
{\displaystyle {\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}}\,{\rm {erf}}\,a{\sqrt {t}}.}
Thus we may write the formula
L
{
e
a
2
t
e
r
f
a
t
}
=
a
(
s
−
a
2
)
s
(
s
>
a
2
)
.
{\displaystyle {\begin{matrix}{\mathcal {L}}\{e^{a^{2}t}\,{\rm {erf}}\,a{\sqrt {t}}\}\;=\;{\frac {a}{(s\!-\!a^{2}){\sqrt {s}}}}\qquad (s>a^{2}).\end{matrix}}}
Moreover, we obtain
1
(
s
+
a
)
s
=
s
−
a
(
s
−
a
2
)
s
=
1
s
−
a
2
−
a
(
s
−
a
2
)
s
↷
e
a
2
t
−
e
a
2
t
e
r
f
a
t
=
e
a
2
t
(
1
−
e
r
f
a
t
)
,
{\displaystyle {\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}\,{\rm {erf}}\,a{\sqrt {t}}\;=\;e^{a^{2}t}(1-{\rm {erf}}\,a{\sqrt {t}}),}
L
{
e
a
2
t
e
r
f
c
a
t
}
=
1
(
a
+
s
)
s
.
{\displaystyle {\begin{matrix}{\mathcal {L}}\{e^{a^{2}t}\,{\rm {erfc}}\,a{\sqrt {t}}\}\;=\;{\frac {1}{(a\!+\!{\sqrt {s}}){\sqrt {s}}}}.\end{matrix}}}
An improper integral
edit
One can utilise the formula (3) for evaluating the improper integral
∫
0
∞
e
−
x
2
a
2
+
x
2
d
x
.
{\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2}}}{a^{2}\!+\!x^{2}}}\,dx.}
e
−
t
x
2
↶
1
s
+
x
2
{\displaystyle e^{-tx^{2}}\;\curvearrowleft \;{\frac {1}{s\!+\!x^{2}}}}
table of Laplace transforms ).\, Dividing this by
a
2
+
x
2
{\displaystyle a^{2}\!+\!x^{2}}
∞
{\displaystyle \infty }
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,
∫
0
∞
e
−
t
x
2
a
2
+
x
2
d
x
=
π
2
a
e
a
2
t
e
r
f
c
a
t
,
{\displaystyle \int _{0}^{\infty }{\frac {e^{-tx^{2}}}{a^{2}\!+\!x^{2}}}\,dx\;=\;{\frac {\pi }{2a}}e^{a^{2}t}\,{\rm {erfc}}\,a{\sqrt {t}},}
∫
0
∞
e
−
x
2
a
2
+
x
2
d
x
=
π
2
a
e
a
2
e
r
f
c
a
.
{\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2}}}{a^{2}\!+\!x^{2}}}\,dx\;=\;{\frac {\pi }{2a}}e^{a^{2}}\,{\rm {erfc}}\,a.}
