 Solution

Take the branch of the complex logarithm with $-\pi <\arg s<\pi$ and write $s^{-1/2}=\exp [-\log s/2]$. Deform the Bromwich contour into a Hankel contour about the negative real axis. The small circle at zero contributes $O(r^{1/2})$ and vanishes; the large closing arcs vanish by the exponential estimates for $t>0$. On the upper lip $s=-r+i0$, $s^{-1/2}=-i/\sqrt{r}$, while on the lower lip it is $i/\sqrt{r}$. Keeping the lip orientations in the Bromwich inversion with a square-root branch cut givesWith $r=u^2$, the supplied Gaussian integral evaluates this as $2{\pi }^{-1}{\int }_0^{\infty }{e}^{-t{u}^2}du$. HenceThe inverse is singular but locally integrable at zero, which is compatible with existence of its Laplace transform.