Can you please show me the solution for the Laplace transform of
$ \frac {e^\frac{-a^2}{4t}}{\sqrt {\pi t}} $
using the definition of Laplace.
The answer must be:
$ \frac {e^\frac {a}{\sqrt s}}{\sqrt s} $
And also what properties of Laplace must be used in solving this type of problem. Thank you.
$$F(s)=\int_0^\infty \mathrm e^{-st}\frac {\mathrm e^\frac{-a^2}{4t}}{\sqrt {\pi t}}\,\mathrm d t$$
Set $u^2=st$, and then $\frac{2}{\sqrt s}\mathrm du=\frac{1}{\sqrt t}\mathrm dt$, so the integral becomes $$ F(s)=\int_0^\infty \mathrm e^{-u^2}\frac {\mathrm e^\frac{-a^2s }{4u^2}}{\sqrt {\pi s}}\,2\,\mathrm d u=\frac{2}{\sqrt {\pi s}}\times\int_0^\infty \mathrm e^{-u^2-\frac{\beta }{u^2}}\,\mathrm d u=\frac{2}{\sqrt {\pi s}}\times \frac12\sqrt\pi\mathrm e^{-2\sqrt\beta} $$ where $\frac{a^2s}{4}=\beta$. So We find