Bromwich integral of an inverse square root (${\int }_{\sigma -i\infty }^{\sigma +i\infty }{e}^{tz}{z}^{-1/2}dz=2i\sqrt{\pi /t}$)

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.