How to obtain inverse Laplace transformation of the following function?

Question

I would like to obtain inverse Laplace transform of a function that includes $(\sqrt{s})$ given inverse Laplace transform of $g(s)$ is obtainable.

$$\mathscr{L}^{-1}\{\frac{g(\sqrt{s})}{s}\}$$

Answer 1

Consider $F(t)=\int_{0}^{\infty}\frac{1}{\sqrt{\pi t}}e^{-u^2/4t}f(u)du$. Now Laplace transform this quantity and exchange the order of integrals to obtain

$$\mathcal{L}F(s)=\int_{0}^{\infty}du~f(u)\int_0^{\infty}\frac{dt}{\sqrt{\pi t}}e^{-st}e^{-\frac{u^2}{4t}}$$

Now isolate the integral with respect to the variable $t$. Take it's Fourier transform and exchange the order of integrals as follows;

$$\mathcal{F}[\int_{0}^{\infty}\frac{dt}{\sqrt{\pi t}}e^{-st}e^{-\frac{u^2}{4t}}](s,\omega)=\int_{0}^{\infty}\frac{dt}{\sqrt{\pi t}}e^{-st}\int_{-\infty}^{\infty}e^{-\frac{u^2}{4t}}e^{i\omega u}du=\int_{0}^{\infty}\frac{dt}{\sqrt{\pi t}}e^{-st}\sqrt{4\pi t}~e^{-\omega^2t}=\frac{2}{\omega^2+s}$$

Inverting the Fourier transform andf using a standard complex residues evaluation we obtain:

$$\int_{0}^{\infty}\frac{dt}{\sqrt{\pi t}}e^{-st}e^{-\frac{u^2}{4t}}=\int_{-\infty}^{\infty}\frac{d\omega}{\pi}\frac{e^{-i\omega u}}{\omega^2+s}=-2\pi i\text{Res}(\frac{e^{-izu}}{z^2+s})\Big|_{z=-i\sqrt{s}}=\frac{e^{-u\sqrt{s}}}{\sqrt{s}}$$

and finally

$$\mathcal{L}F(s)=\int_0^{\infty}du~f(u)\frac{e^{-u\sqrt{s}}}{\sqrt{s}}=\frac{\mathcal{L}f(\sqrt{s})}{\sqrt{s}}$$

which in the OP's notation:

$$\mathcal{L}^{-1}\Big[\frac{g(\sqrt{s})}{\sqrt{s}}\Big](t)=F(t)$$

and evidently the statement in the book is false. However, the statement can be saved by a "small" modification:

$$\mathcal{L}^{-1}[\frac{g(\sqrt{s})}{s}](t)=\int_0^{\infty}du~f(u)\mathcal{L}^{-1}[\frac{e^{-u\sqrt{s}}}{s}](t)=\int_{0}^{\infty}du~f(u)\text{erfc}(\frac{u}{2\sqrt{t}})$$

Note: The computation of the above inverse LT is done in Schaum p.207