Contour Integral of Gamma Function

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Theorem

Let $\Gamma$ 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:

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

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:

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

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

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

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

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

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:

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

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

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

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:

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

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:

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

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

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

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

Thus we have:

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

Which gives us:

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

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

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

$\blacksquare$