# 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 $$\delta>0$$

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 :)