I recently read a proof for the following formula which I don't understand completely. For $Re(z)>0$:
$\Gamma(z)=\frac{1}{e^{2\pi iz}-1}\int_{C_\delta}e^{-\zeta}\zeta^{z-1}d\zeta$ , where $C_\delta$ is the $\delta$-Hankel Contour.
The proof:
Let $\delta>0$ be arbitrary. Then we have: $$\begin{align} G(z)&:=\int_{C_\delta}e^{-\zeta}\zeta^{z-1}d\zeta \\ &=-\int_\delta^\infty e^{-t}t^{z-1}dt+\int_\delta^\infty e^{-t}t^{(z-1)(\ln t+2\pi i)}dt+\int_{\partial B(0,\delta)}e^{-\zeta}\zeta^ {z-1}d\zeta \\ &=(e^{2\pi i}-1)\int_\delta^\infty e^{-t}t^{z-1}dt+\int_{\partial B(0,\delta)}e^{-\zeta}\zeta^{z-1}d\zeta \end{align}$$ For some $z$-dependant constant $\alpha_z>0$, we have the following inequality. $$ \left|\int_{\partial B(0,\delta)}e^{-\zeta}\zeta^{z-1}d\zeta\right|\leq 2\pi\delta^{Re(z)}\alpha_z $$
For $Re(z)>0, \delta\to0$ implies:$\int_{\partial B(0,\delta)}e^{-\zeta}\zeta^{z-1}d\zeta\to 0$.
Therefore we have: $$ G(z)=(e^{2\pi i}-1)\Gamma(z) $$
The first line doesn't make sense to me since it is using two different definitions of the function $\zeta^{z-1}$. I also don't understand why the integral is independent of the choice of $\delta>0$