Inverse Laplace Transform and error function

Question

Express your answer in terms of the error function:

$$L^{-1}\left[\frac{1}{\sqrt{s^3+as^2}}\right]$$

Clue: $\qquad L\left[\frac{1}{\sqrt{t}}\right]=\sqrt\frac{π}{s} \qquad , \qquad s>0$

Error function: $\frac{2}{\sqrt{π}}\int_0^te^{-w^2}dw$

Answer 1

The answer, of course, is

$$\mathcal{L}^{-1} \left (\frac1{s \sqrt{s+a}} \right ) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{s+a}} = \frac{\operatorname{erf}{\sqrt{a t}}}{\sqrt{a}} $$

Convolution

Note that

$$\mathcal{L} \left (t^{-1/2} \right ) = \sqrt{\pi} s^{-1/2} \implies \mathcal{L} \left (t^{-1/2} e^{-a t} \right ) = \sqrt{\pi} (s+a)^{-1/2}$$

or, written another way

$$\mathcal{L}^{-1} \left (\frac1{\sqrt{s+a}} \right ) = (\pi t)^{-1/2} e^{-a t}$$

Thus, by the convolution theorem,

$$\mathcal{L}^{-1} \left (\frac1{s \sqrt{s+a}} \right ) = \pi^{-1/2} \int_0^t dt' \, t'^{-1/2} \, e^{-a t'}= \frac1{\sqrt{a}} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{a t}} du \, e^{-u^2}$$

The result follows.