I am trying to show that the Laplace transform of erf(ā(t)) is equal to 1/(sā(s+1))
I have started with the definition of erf(t) as (2/āš)times the integral from t-0 of e^(-x^2) and substituted ā(t) for t. I then used a Taylor series for e^(-x^2) and took the Laplace transform of that. I'm left with a series I don't know what to do with as I don't see a direct route to what I'm trying to get to: 1/(sā(s+1)). Am I even on the right track? Is there an easier way to do this?
Best Answer
Let $f(s)$ be the Laplace transform of erf$(\sqrt{t})$:
$$f(s)=\frac{2}{\sqrt{\pi}}\int_0^{+\infty}dt\,e^{-st}\int_0^\sqrt{t}dx\,e^{-x^2}=\frac{2}{\sqrt{\pi}}\int\int_\Omega dtdx\, e^{-x^2-st}\,,$$
where $\Omega$ is the region in the first quadrant of the $x$-$t$ plane bounded by the curve $t=x^2$ and the (vertical) $t$ axis. Interchanging the order of the integrals, i.e. doing first the integral in $t$ and then the integral in $x$, one gets:
$$f(s)=\frac{2}{\sqrt{\pi}}\int_0^{+\infty}dx\,e^{-x^2}\int_{x^2}^{+\infty}dt\,e^{-st} = \frac{2}{s\sqrt{\pi}}\int_0^{+\infty}dx\,e^{-(s+1)x^2} = \frac{1}{s\sqrt{s+1}}\,.$$