Bromwich inversion with a square-root branch cut

When an inverse Laplace integrand contains $p^{-1/2}$ on the principal branch, closing the Bromwich contour to the left encloses isolated poles and wraps a branch cut along the negative real axis. Pole residues give persistent oscillatory terms, while the jump across the two sides of the cut gives a real decaying integral.

For $\sigma ,t>0$, take the principal square root and orient the Bromwich contour upward. Closing left around the negative-real branch cut gives the jump $2i{\int }_0^{\infty }{e}^{-tx}{x}^{-1/2}dx$. The substitution $u=tx$ and $\Gamma (1/2)=\sqrt{\pi }$ prove the displayed formula. A rectangular closure has horizontal remainders $O(R^{-1/2})$, an exponentially small left edge, and an $O(\sqrt{\epsilon })$ small-circle term. These estimates also justify the conditionally convergent vertical integral.