= Bromwich integral of an inverse square root
{c}
{title2=$\int_{\sigma-i\infty}^{\sigma+i\infty}e^{tz}z^{-1/2}dz=2i\sqrt{\pi/t}$}
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 $O(\sqrt\varepsilon)$ small-circle term. These estimates also justify the conditionally convergent vertical <integral>.
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