Contour Integral of Gamma Function

Theorem
Let $\Gamma \left({z}\right)$ denote the gamma function.

Let $y$ be a positive number.

Then for any positive number $c$:


 * $\displaystyle \frac 1 {2 \pi i} \int_{c - i \infty}^{c + i \infty} \Gamma \left({t}\right) y^{-t} \rd t = e^{-y}$

Proof
Let $L$ be the rectangular contour with the vertices $c \pm iR$, $-N - \frac 1 2 \pm iR$.

We will take the Contour Integral of $\Gamma \left({t}\right) y^{-t}$ about the rectangular contour, $L$.

Note from Poles of Gamma Function, that the poles of this function are located at the non-positive integers.

Thus, by the Residue Theorem, we have:


 * $\displaystyle \oint_L \Gamma \left({t}\right) y^{-t} \, \mathrm d z = 2 \pi i \displaystyle\sum_{k\mathop = 0}^N \operatorname{Res} \left({-k}\right)$

Thus, we obtain:


 * $\displaystyle \lim_{N \to \infty} \displaystyle \lim_{R \to \infty} \displaystyle \oint_L \Gamma \left({t}\right) y^{-t} \mathrm d z = 2 \pi i\displaystyle \sum_{k\mathop = 0}^{\infty} \operatorname{Res} \left({-k}\right)$

From Residues of Gamma Function, we see that:


 * $\operatorname{Res} \left({-k}\right) = \dfrac{ \left({-1}\right)^k y^{k} }{k!} $

Which gives us:

We aim to show that the all but the righthand side of the rectangular contour go to 0 as we take these limits, as our result follows readily from this.

The top and bottom portions of the contour can be parameterized by:


 * $\gamma\left({t}\right)= c\pm iR - t$

where $0 < t < c + N + \frac 1 2$.

The modulus of the contour integral is therefore given by:

From Bound on Complex Values of Gamma Function, we have that:

for all $|R|\geq 1$. Because $|R|\geq 1$, we have that


 * Combining the two inequalities we obtain:

We see that

as the poles of Gamma are at the nonpositive integers, which means that the integral is a definite integral of a continuous function. The above is enough to allow for the interchange of limits by the Dominated Convergence Theorem, thus we have

But using Equation $(1)$ from above we see:

Thus by the Squeeze Theorem we have
 * $ \displaystyle \lim_{R \to \infty} \left\vert \Gamma \left({c\pm iR - t}\right) \right\vert = 0$

Which means we have

Thus we have that the top and bottom of the contour go to $0$ in the limit.

The of the contour may be parameterized by:


 * $\gamma\left(t\right)= -N-\dfrac 1 2 -it$

where $t$ runs from $-R$ to $R$.

Thus the absolute value of integral of the left-hand side is given as:

Thus we have

Which gives us

Thus we have the left, top, and bottom of the rectangular contour go to 0 in the limit, which gives us