# Contour Integral of Gamma Function

## Theorem

Let $\Gamma$ denote the gamma function.

Let $y$ be a (strictly) positive real number.

Then for any (strictly) positive real 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:

 $\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \oint_L \map \Gamma t y^{-t} \rd t$ $=$ $\ds 2 \pi i \sum_{k \mathop = 0}^\infty \map {\operatorname{Res} } {-k}$ $\ds$ $=$ $\ds 2 \pi i \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k y^k} {k!}$ $\ds$ $=$ $\ds 2 \pi i e^{-y}$ Power Series Expansion for Exponential Function

We aim to show that the all but the right hand 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:

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

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

The modulus of the contour integral is therefore given by:

 $\ds \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {\map \gamma t} y^{- \map \gamma t} \map {\gamma'} t \rd t}$ $=$ $\ds \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^{-\paren {c \pm i R - t} } \rd t}$ $\ds$ $=$ $\ds \cmod {y^{-\paren {c \pm i R} } } \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^t \rd t}$ $\ds$ $=$ $\ds y^{-c} \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^t \rd t}$

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

 $\ds \cmod {\map \Gamma {c \pm i R - t} y^t}$ $\le$ $\ds \frac {\cmod {c - t + i} } {\cmod {c - t + i R} } \cmod {\map \Gamma {c - t + i} y^t}$ (1)

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

 $\ds \frac {\cmod {c - t + i} } {\cmod {c - t + i R} }$ $\le$ $\ds 1$
Combining the two inequalities we obtain:
 $\ds \cmod {\map \Gamma {c \pm i R - t} y^t}$ $\le$ $\ds \cmod {\map \Gamma {c - t + i} y^t}$ (2)

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 non-positive 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:

 $\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} y^{-c} \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^t \rd t}$ $=$ $\ds \lim_{N \mathop \to \infty} y^{-c} \cmod {\int_0^{c + N + \frac 1 2} \lim_{R \mathop \to \infty} \map \Gamma {c \pm i R - t} y^t \rd t}$

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

 $\ds 0$ $\le$ $\ds \lim_{R \mathop \to \infty} \cmod {\map \Gamma {c \pm i R - t} y^t}$ $\ds$ $\le$ $\ds \lim_{R \mathop \to \infty} \frac {\cmod {c - t + i} } {\cmod {c - t + i R} } \cmod {\map \Gamma {c - t + i} y^t}$ $\ds$ $=$ $\ds 0$

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:

 $\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} y^{-c} \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^t \rd t}$ $=$ $\ds \lim_{N \mathop \to \infty} y^{-c} \cmod {\int_0^{c + N + \frac 1 2} \lim_{R \mathop \to \infty} \map \Gamma {c \pm i R - t} y^t \rd t}$ $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} y^{-c} \cmod {\int_0^{c + N + \frac 1 2} 0 y^t \rd t}$ $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} 0$ $\ds$ $=$ $\ds 0$

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

The left hand side 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 left hand side is given as:

 $\ds \cmod {\int_{-R}^R \map \Gamma {\map \gamma t} y^{- \map \gamma t} \map {\gamma'} t \rd t}$ $=$ $\ds \cmod {\int_{-R}^R \map \Gamma {- N - \frac 1 2 - i t} y^{- \paren {- N - \frac 1 2 - i t} } \paren {-i} \rd t}$ $\ds$ $=$ $\ds \cmod {\int_{-R}^R \frac {\map \Gamma {- N + \frac 3 2 - i t} } {\paren {- N - \frac 1 2 - i t} \paren {- N + \frac 1 2 - i t} } y^{- \paren {- N - \frac 1 2 - i t} } \paren {-i} \rd t}$ Gamma Difference Equation $\ds$ $\le$ $\ds \int_{-R}^R \cmod {\frac {\map \Gamma {- N + \frac 3 2 - i t} } {\paren {- N - \frac 1 2 - i t} \paren {- N + \frac 1 2 - i t} } y^{- \paren {- N - \frac 1 2 - i t} } \paren {-i} } \rd t$ Modulus of Complex Integral $\ds$ $=$ $\ds \int_{-R}^R \cmod {\frac {\map \Gamma {- N + \frac 3 2 - i t} } {\paren {- N - \frac 1 2 - i t} \paren {- N + \frac 1 2 - i t} } y^{N + \frac 1 2} } \rd t$ $\ds$ $\le$ $\ds \int_{-R}^R \frac {\cmod {\map \Gamma {- N + \frac 3 2} } } {\cmod {\paren {- N - \frac 1 2 -i t} \paren {- N + \frac 1 2 - i t} } } y^{N + \frac 1 2} \rd t$ See equation (2) above $\ds$ $=$ $\ds \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \int_{-R}^R \frac 1 {\cmod {\paren {- N - \frac 1 2 -i t} \paren {- N + \frac 1 2 - i t} } } \rd t$ $\ds$ $\le$ $\ds \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \int_{-R}^R \frac 1 {\cmod {\paren {- N + \frac 1 2 - i t} }^2} \rd t$ $\ds$ $=$ $\ds \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \int_{-R}^R \frac 1 {\paren {- N + \frac 1 2}^2 + t^2} \rd t$ Definition of Complex Modulus $\ds$ $=$ $\ds \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \frac {\map \arctan {\frac R {- N + \frac 1 2} } - \map \arctan {\frac {-R} {- N + \frac 1 2} } }{- N + \frac 1 2}$ Derivative of Arctangent Function/Corollary

Thus we have:

 $\ds 0$ $\le$ $\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \cmod {\int_{-R}^R \map \Gamma {- N - \frac 1 2 - i t} y^{- \paren {- N - \frac 1 2 - i t} } \paren {-i} \rd t}$ $\ds$ $\le$ $\ds \lim_{N \mathop \to \infty} \lim_{R \to \infty} \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \frac {\map \arctan {\frac R {-N+ \frac 1 2} } - \map \arctan {\frac {-R} {- N + \frac 1 2} } } {- N + \frac 1 2}$ $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \frac \pi {- N + \frac 1 2}$ $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} \frac {2^{2 N - 2} \paren {N - 1}!} {\paren {2 N - 2}!} \sqrt \pi y^{N + \frac 1 2} \frac \pi {- N + \frac 1 2}$ Gamma Function of Negative Half-Integer $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} \frac{2^{2 N - 2} \sqrt {2 \pi \paren {N - 1} } \paren {\frac {N - 1} e}^{N - 1} } {\sqrt{2 \pi \paren {2 N - 2} } \paren {\frac {2 \paren {N - 1} } e}^{2 N - 2} } \sqrt \pi y^{N + \frac 1 2} \frac \pi {- N + \frac 1 2}$ Stirling's Formula $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} \frac{2^{2 N - 2} \paren {\frac {N - 1} e}^{N - 1} } {2^{2 N -2} \sqrt{2} \paren {\frac {N - 1} e}^{2 N - 2} } \sqrt \pi y^{N + \frac 1 2} \frac \pi {-N + \frac 1 2}$ $\ds$ $=$ $\ds \lim_{N \mathop \to \infty} \frac 1 {\sqrt 2 \paren {\frac {N - 1} e}^{N - 1} } \sqrt \pi y^{N + \frac 1 2} \frac \pi {- N + \frac 1 2}$ $\ds$ $=$ $\ds 0$

which gives us:

 $\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \cmod {\int_{-R}^R \map \Gamma {- N - \frac 1 2 - i t} y^{- \paren {- N - \frac 1 2 - i t} } \paren {-i} \rd t}$ $=$ $\ds 0$ Squeeze Theorem for Functions

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

 $\ds \frac 1 {2 \pi i} \int_{c - i \infty}^{c + i \infty} \map \Gamma t y^{-t} \rd t$ $=$ $\ds \frac 1 {2 \pi i} \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \oint_L \map \Gamma t y^{-t} \rd t$ $\ds$ $=$ $\ds e^{-y}$

$\blacksquare$