Contour Integral of Gamma Function
In particular: throughout.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this in more detail, feel free to use the talk page.
Once you have done so, you can remove this instance of {{MissingLinks}} from the code.
throughout
It may also need to be brought up to our standard house style.
If you have done this or know it has been done, please remove {{Tidy}} from the code.
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.
This should be made clearer
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
The top and bottom portions of the contour can be parameterized by:
- $\gamma\left({t}\right)= c\pm iR - t$
where $0 < t < c + N + \frac 1 2$.
A labelled diagram here would help
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
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.
I know you're referring to convergence here, but you should make it clearer.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
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:
| \(\displaystyle \dfrac{1}{2\pi i} \displaystyle \int_{c-i\infty}^{c+i\infty} \Gamma \left({t}\right) y^{-t} \rd t\) | \(=\) | \(\displaystyle \dfrac{1}{2\pi i} \lim_{N \to \infty} \lim_{R \to \infty} \oint_L \Gamma \left({t}\right) y^{-t} \, \mathrm d t\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle e^{-y}\) | $\quad$ | $\quad$ |
$\blacksquare$
