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 ?
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