Hankel curve formula for gamma function proof 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 $\delta>0$
 A: Here's a sketch of a slightly different approach that you might find helpful.
Consider $f(z, t)=t^{z-1}e^{-t}$ and its $t$-dependence. We need to define the power function $t^{z-1}$ for complex $z$ as a single-valued function, so let's make a cut $[0, \infty)$ in the extended complex plane $\tilde{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}$ and note that the Hankel contour surrounds that cut. Write $x+iy=t=re^{i\theta}$ for $\theta\in(0, 2\pi)$. Then we can take $$t^{z-1}=r^{z-1}e^{i\theta (z-1)}.$$
Now look at $f$ above and below the cut (so $x>0$): $$\begin{align}f(z, x+i0)&=x^{z-1}e^{-x} \\
f(z, x-i0)&=e^{2\pi iz}x^{z-1}e^{-x}=e^{2\pi iz}f(z, x+i0). \end{align}$$
But $f$ is analytic everywhere outside the cut, so that $I(z):=\int_{C_\delta}f(z, t)\, dt$ is defined for all $z\in\mathbb{C}$ as there are no singularities on the contour (and the integral converges due to $e^{-t}$).
Assume $\operatorname{Re}(z)>0$ so that the Euler integral definition of $\Gamma$ can be used. Then we can decompose $$I(z)=\int_\delta^\infty f(z, x-i0)\, dt+\int_{\infty}^{\delta}f(z, x+i0) + \underbrace{\int_{\partial B(0, \delta )}f(z, t)\, dt}_{(*)}\tag{1}$$ so that $(*)$ vanishes as $\delta\to0$. [The independence of the choice of $\delta$ follows from standard results in complex analysis; viz., that one can deform a contour within some open set without changing the integral over that contour, so long as one is careful to avoid singularities.]
But now $(1)$ can be rewritten to give the desired result (can you see why?), which, by the principle of analytical continuation, can be extended to $\mathbb{C}\backslash\{0, -1, -2, \dots \}$ since $I(z)$ is analytic for all $z\in\mathbb{C}$, and for $n\in\mathbb{N}$, the integrand $f(n, t)$ is analytic for all $t\in\mathbb{C}$ (meaning the cut disappears and $I(n)=0$, cancelling the zeros of the denominator).

Phew! That took longer to type up than I thought it would! I hope that helps :)
