# 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:

 $\displaystyle \lim_{N \to \infty} \lim_{R \to \infty} \oint_L \Gamma \left({t}\right) y^{-t} \, \mathrm d t$ $=$ $\displaystyle 2 \pi i\displaystyle \sum_{k\mathop = 0}^{\infty} \operatorname{Res} \left({-k}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 2 \pi i \sum_{k\mathop = 0}^{\infty} \frac{ \left({-1}\right)^k y^{k} }{k!}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 2 \pi i e^{-y}$ $\quad$ Power Series Expansion for Exponential Function $\quad$

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:

 $\displaystyle \left\vert \int_0^{c +N + \frac 1 2} \Gamma \left({\gamma\left(t\right)}\right) y^{-\gamma\left(t\right)} \gamma'\left(t\right) \rd t \right\vert$ $=$ $\displaystyle \left\vert \int_0^{c +N + \frac 1 2} \Gamma \left({c\pm iR - t}\right) y^{-\left(c\pm iR - t\right)} \rd t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left\vert y^{-\left({c\pm iR}\right)} \right\vert \left\vert \int_0^{c + N + \frac 1 2} \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle y^{-c} \left\vert \int_0^{c + N + \frac 1 2} \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert$ $\quad$ $\quad$

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

 $\displaystyle \left\vert \Gamma \left({c\pm iR - t}\right) y^t \right\vert$ $\leq$ $\displaystyle \dfrac{\left\vert c-t+i \right\vert } {\left\vert c-t+iR \right\vert}\left\vert \Gamma \left({c-t+i}\right) y^t \right\vert$ $\quad$ (1) $\quad$

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

 $\displaystyle \frac{\left\vert c-t+i\right\vert}{\left\vert c-t+iR \right\vert}$ $\leq$ $\displaystyle 1$ $\quad$ $\quad$
Combining the two inequalities we obtain:
 $\displaystyle \left\vert \Gamma \left({c\pm iR - t}\right) y^t \right\vert$ $\leq$ $\displaystyle \left\vert \Gamma \left({c-t+i}\right) y^t \right\vert$ $\quad$ (2) $\quad$

We see that:

$\displaystyle \int_0^{c + N + \frac 1 2 } \left\vert \Gamma \left({c- t+i}\right) y^t \right\vert \rd t < \infty$

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:

 $\displaystyle \lim_{N \to \infty} \lim_{R \to \infty} y^{-c} \left\vert \displaystyle \int_0^{c+N+\frac{1}{2} } \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert$ $=$ $\displaystyle \lim_{N \to \infty} y^{-c} \left\vert \displaystyle \int_0^{c+N+\frac{1}{2} } \lim_{R \to \infty} \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert$ $\quad$ $\quad$

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

 $\displaystyle 0$ $\leq$ $\displaystyle \lim_{R \to \infty} \left\vert \Gamma \left({c\pm iR - t}\right) y^t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $\leq$ $\displaystyle \lim_{R \to \infty} \dfrac{\left\vert c-t+i \right\vert } {\left\vert c-t+iR \right\vert}\left\vert \Gamma \left({c-t+i}\right) y^t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

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:

 $\displaystyle \lim_{N \to \infty} \lim_{R \to \infty} y^{-c} \left\vert \displaystyle \int_0^{c+N+\frac{1}{2} } \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert$ $=$ $\displaystyle \lim_{N \to \infty} y^{-c} \left\vert \displaystyle \int_0^{c+N+\frac{1}{2} } \lim_{R \to \infty} \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} y^{-c} \left\vert \displaystyle \int_0^{c+N+\frac{1}{2} } 0 y^t \rd t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} 0$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

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:

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

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

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

 $\displaystyle \left\vert \int_{-R}^R \Gamma \left({\gamma\left(t\right)}\right) y^{-\gamma\left(t\right)} \gamma'\left(t\right) \rd t \right\vert$ $=$ $\displaystyle \left\vert \int_{-R}^R \Gamma \left({-N-\dfrac 1 2 -it}\right) y^{-\left(-N-\dfrac 1 2 -it\right)} \left( -i \right) \rd t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left\vert \int_{-R}^R \dfrac{ \Gamma \left({-N+\dfrac 3 2 -it}\right)}{\left(-N-\dfrac 1 2 -it\right)\left(-N+\dfrac 1 2-it\right)} y^{-\left(-N-\dfrac 1 2 -it\right)} \left( -i \right) \rd t \right\vert$ $\quad$ Gamma Difference Equation $\quad$ $\displaystyle$ $\leq$ $\displaystyle \int_{-R}^R \left\vert \dfrac{ \Gamma \left({-N+\dfrac 3 2 -it}\right)}{\left(-N-\dfrac 1 2 -it\right)\left(-N+\dfrac 1 2-it\right)} y^{-\left(-N-\dfrac 1 2 -it\right)} \left( -i \right) \right\vert \rd t$ $\quad$ Modulus of Complex Integral $\quad$ $\displaystyle$ $=$ $\displaystyle \int_{-R}^R \left\vert \dfrac{ \Gamma \left({-N+\dfrac 3 2 -it}\right)}{\left(-N-\dfrac 1 2 -it\right)\left(-N+\dfrac 1 2-it\right)} y^{N+\dfrac 1 2} \right\vert \rd t$ $\quad$ $\quad$ $\displaystyle$ $\leq$ $\displaystyle \int_{-R}^R \dfrac{\left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert}{\left\vert \left(-N-\dfrac 1 2 -it\right)\left(-N+\dfrac 1 2-it\right)\right\vert} y^{N+\dfrac 1 2} \rd t$ $\quad$ See equation (2) above $\quad$ $\displaystyle$ $=$ $\displaystyle \left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert y^{N+\dfrac 1 2} \int_{-R}^R \dfrac 1 {\left\vert \left(-N-\dfrac 1 2 -it\right)\left(-N+\dfrac 1 2-it\right)\right\vert} \rd t$ $\quad$ $\quad$ $\displaystyle$ $\leq$ $\displaystyle \left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert y^{N+\dfrac 1 2} \int_{-R}^R \dfrac 1 {\left\vert\left(-N+\dfrac 1 2-it\right)\right\vert^2} \rd t$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert y^{N+\dfrac 1 2} \int_{-R}^R \dfrac 1 {\left(-N+\dfrac 1 2\right)^2+t^2} \rd t$ $\quad$ Definition of Complex Modulus $\quad$ $\displaystyle$ $=$ $\displaystyle \left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert y^{N+\dfrac 1 2} \dfrac {\arctan\left(\dfrac{R}{-N+ \frac 1 2}\right) - \arctan\left(\dfrac{-R}{-N+ \frac 1 2}\right)}{-N+\frac 1 2}$ $\quad$ Derivative of Arctangent Function/Corollary $\quad$

Thus we have:

 $\displaystyle 0$ $\leq$ $\displaystyle \lim_{N \to \infty} \lim_{R \to \infty} \left\vert \displaystyle \int_{-R}^{R } \Gamma \left({-N-\dfrac 1 2 -it}\right) y^{-\left(-N-\dfrac 1 2 -it\right)} \left(-i\right) \rd t \right\vert$ $\quad$ $\quad$ $\displaystyle$ $\leq$ $\displaystyle \lim_{N \to \infty} \lim_{R \to \infty} \left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert y^{N+\dfrac 1 2} \dfrac {\arctan\left(\dfrac{R}{-N+ \frac 1 2}\right) - \arctan\left(\dfrac{-R}{-N+ \frac 1 2}\right)}{-N+\frac 1 2}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} \left\vert \Gamma \left({-N+\dfrac 3 2}\right)\right\vert y^{N+\dfrac 1 2} \dfrac {\pi}{-N+\frac 1 2}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} \frac {2^{2N-2} \left(N-1\right)!} {\left({2N-2}\right)!} \sqrt \pi y^{N+\dfrac 1 2} \dfrac {\pi}{-N+\frac 1 2}$ $\quad$ Gamma Function of Negative Half-Integer $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} \frac{2^{2N-2} \sqrt {2 \pi \left(N-1\right)} \left({\dfrac {N-1}{e} }\right)^{N-1} } {\sqrt{2 \pi \left(2N-2\right)} \left({\dfrac {2\left(N-1\right)}{e} }\right)^{2N-2} } \sqrt \pi y^{N+\dfrac 1 2} \dfrac {\pi}{-N+\frac 1 2}$ $\quad$ Stirling's Formula $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} \frac{2^{2N-2} \left({\dfrac {N-1}{e} }\right)^{N-1} } { 2^{2N-2} \sqrt{2} \left({\dfrac {N-1}{e} }\right)^{2N-2} } \sqrt \pi y^{N+\dfrac 1 2} \dfrac {\pi}{-N+\frac 1 2}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{N \to \infty} \frac{1} {\sqrt{2} \left({\dfrac {N-1}{e} }\right)^{N-1} } \sqrt \pi y^{N+\dfrac 1 2} \dfrac {\pi}{-N+\frac 1 2}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

Which gives us:

 $\displaystyle \lim_{N \to \infty} \lim_{R \to \infty} \left\vert \displaystyle \int_{-R}^{R } \Gamma \left({-N-\dfrac 1 2 -it}\right) y^{-\left(-N-\dfrac 1 2 -it\right)} \left( -i \right) \rd t \right\vert$ $=$ $\displaystyle 0$ $\quad$ Squeeze Theorem $\quad$

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

 $\displaystyle \dfrac{1}{2\pi i} \displaystyle \int_{c-i\infty}^{c+i\infty} \Gamma \left({t}\right) y^{-t} \rd t$ $=$ $\displaystyle \dfrac{1}{2\pi i} \lim_{N \to \infty} \lim_{R \to \infty} \oint_L \Gamma \left({t}\right) y^{-t} \, \mathrm d t$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle e^{-y}$ $\quad$ $\quad$

$\blacksquare$