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