How to derive this Hankel's Contour integral formula with gamma function? This relation was put up in The Art Of Computer Programming and no derivation was offered. Please help me understand this better.
$$\frac{1}{\Gamma (z)} = \frac{1}{2i\pi} \oint\frac{e^t dt}{t^z}$$
It said that the path of the complex integration starts at $-\infty$ circles around the origin and returns to $-\infty$. 
If not a derivation, at least help me develop some intuition about this. How are we introducing complex analysis to a function that came up in the real numbers ? 
 A: Ok, i don't want to be a jerk, so here we go. Let's define the complex valued function 
$$
f(z)=\frac{e^{z}}{z^x}
$$
We integrate $f(z)$ over the so called Hankel contour $\mathcal{H}$ which consists of three parts ($\delta \rightarrow 0_+$):
-the line segment $[-\infty+i\delta,i\delta]$, denoted by $l_+$
-a small semicircle around the origin with radius $\delta$, denoted by $sc$
-the line segment $[-\infty-i\delta,-i\delta]$, denoted by $l_-$
Note that this contour fixes the branch cut of log to lie on the negative real axis.
It is not difficult to show that $\int_{sc}f(z)dz$ yields a zero contribution as long as $z>-1$ so we end up,using Cauchy's integral theorem, with
$$
\int_{\mathcal{H}} f(z)dz=-\int_{l+}f(z)dz-\int_{l-}f(z)dz
$$
now noting that $\lim_{\delta\rightarrow 0}f(t\pm i\delta)=e^{\mp i \pi x }\frac{e^{t}}{|t|^x}$ we obtain
$$
\int_{\mathcal{H}}  f(z)dz =2i\sin(\pi x)\int_{-\infty}^0\frac{e^t}{|t|^x}dt=2i\sin(\pi x)\int_0^{\infty}\frac{e^{-t}}{t^x}dt
$$
the last integral is the standard integral representation of the gamma function
$$
\int_{\mathcal{H}}  f(z)dz=2i \sin(\pi x)\Gamma(1-x)
$$
thanks to the well known multiplicative property of the gamma function this equals

$$
\int_{\mathcal{H}}  f(z)dz=\frac{2i\pi}{\Gamma(x)} 
$$
  QED

