I posted some examples before such as
$$\mathscr{L}^{-1}\left[\frac{e^{-\alpha \sqrt{s}}}{\sqrt{s}(s-A)}\right]=\frac{e^{-\alpha \sqrt{A}+At}}{\sqrt{A}} - \frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{1/2}(r+A)}dr$$ and they have been solved by friends here (How to get the inverse Laplace Transform?).
I have been trying $$\frac{e^{-\alpha \sqrt{s}}}{s^{3/2}}$$ by the same way but failed, because the result is $$-\frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{3/2}}dr$$ and it is divergent.
Moreover, $$\lim_{A \to 0}\left[\frac{e^{-\alpha \sqrt{A}+At}}{\sqrt{A}} - \frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{1/2}(r+A)}dr]\right] \nrightarrow -\frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{3/2}}dr$$
I think the branch cut along the minus axis might be wrong.
Would you like to help me?
Mathematica can fulfill this inverse Laplace transform: $$\mathscr{L}^{-1}\left[\frac{e^{-\alpha \sqrt{s}}}{s^{3/2}}\right] = -\alpha + \frac{2\sqrt{t}}{\sqrt{\pi}}e^{-\frac{\alpha^2}{4t}} +\alpha erf\left(\frac{\alpha}{2\sqrt{t}}\right)$$.
But I really want to know this fulfillment by residue theorem.