Contour Integral of Gamma Function

Theorem
Let $\Gamma$ denote the gamma function.

Let $y$ be a positive number.

Then for any positive number $c$:


 * $\ds \frac 1 {2 \pi i} \int_{c - i \infty}^{c + i \infty} \map \Gamma t y^{-t} \rd t = e^{-y}$

Proof
Let $L$ be the rectangular contour with the vertices $c \pm i R$, $- N - \dfrac 1 2 \pm i R$.

We will take the Contour Integral of $\map \Gamma t y^{-t}$ 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 Cauchy's Residue Theorem:


 * $\ds \oint_L \map \Gamma t y^{-t} \rd z = 2 \pi i \sum_{k \mathop = 0}^N \map {\operatorname{Res} } {-k}$

Thus, we obtain:


 * $\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \oint_L \map \Gamma t y^{-t} \rd z = 2 \pi i \sum_{k \mathop = 0}^\infty \map {\operatorname{Res} } {-k}$

From Residues of Gamma Function, we see that:


 * $\map {\operatorname{Res} } {-k} = \dfrac {\paren {-1}^k y^k} {k!}$

Which gives us:

We aim to show that the all but the 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:


 * $\map \gamma t = c \pm i R - t$

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

The modulus of the contour integral is therefore given by:

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

for all $\cmod R \ge 1$. Because $\cmod R \ge 1$, we have that


 * Combining the two inequalities we obtain:

We see that:


 * $\ds \int_0^{c + N + \frac 1 2} \cmod {\map \Gamma {c- t + i} y^t} \rd t < \infty$

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 for Functions we have:


 * $\ds \lim_{R \mathop \to \infty} \cmod {\map \Gamma {c \pm i R - t} } = 0$

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:


 * $\map \gamma t = N - \dfrac 1 2 - it$

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

Thus the absolute value of integral of the 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: