I have a question about the Hankel path representation of the Gamma function. The path of integration is displayed on the imege. The branch cut is taken as the negative real axis. $$ \Gamma(z) = \frac{1}{2 i \sin { \pi z}} \int_{Ha} e^t t^{z-1} dt $$
In my derivation I actually get $$ \Gamma(z) = \frac{-1}{2 i \sin { \pi z}} \int_{Ha} e^t t^{z-1} dt $$ by the substitution along AB: $t= x e^{i\pi}$ and along ED: $ x e^{- i\pi}$. Let $z>0$ so that the arch integral would vanish.
The sum of the ray integrals eventually becomes $$ I = - e^{i \pi z} \int_{\epsilon}^{\infty} e^{-x} x^{z-1} dx + e^{-i \pi z} \int_{\epsilon}^{\infty} e^{-x} x^{z-1} dx $$ which for $\epsilon \rightarrow 0$ becomes $$ I = - 2i\sin{\pi z} \; \Gamma(z) $$
Consider the ray $DE$. The integral is $$\int_{-\infty}^0 (t - i \epsilon)^{z - 1} e^{t - i \epsilon} dt = \int_{-\infty}^0 e^{(z - 1)(\ln |t - i \epsilon| + i \arg(t - i \epsilon))} e^{t - i \epsilon} dt \to \\ e^{-i \pi (z - 1)} \int_{-\infty}^0 (-t)^{z - 1} e^t dt = -e^{-i \pi z} \int_0^\infty t^{z - 1} e^{-t} dt,$$ so the signs in $I$ are off.