Hankel's representation for reciprocal Gamma

First assume $0<\mathrm{Re}z<1$. On the small circle about $0$, the integrand has size $O(r^{-\mathrm{Re}z})$, so the circular contribution tends to $0$ as $r\downarrow 0$. On the two rays, $e^t$ decays exponentially as $t=-x$ with $x\to +\mathrm{\infty }$, so the contour integral converges absolutely.

Along the upper and lower sides of the cut, one has $(-x\pm i0{)}^{-z}=e^{\mp \pi iz}{x}^{-z}$. With the orientation from [L3], the lower ray contributes $e^{\pi iz}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-z}dx$ and the upper ray contributes $-e^{-\pi iz}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-z}dx$. Therefore step 1.1 yields ${\int }_H{e}^t{t}^{-z}dt=(e^{\pi iz}-e^{-\pi iz}){\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-z}dx=2i\sin (\pi z)\mathrm{\Gamma }(1-z)\mathrm{.}$

Using [L1], the right-hand side of step 2.1 equals $2\pi i\mathrm{/}\mathrm{\Gamma }(z)$ on $0<\mathrm{Re}z<1$. Hence the displayed integral formula holds on that strip. Both sides are meromorphic in $z$, and [L2] extends the identity to all of $\mathbb{C}$.