Theorem: Contour Integral of Gamma Function

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: 
{{begin-eqn}}
{{eqn | l = \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \oint_L \map \Gamma t y^{-t} \rd t
      | r = 2 \pi i \sum_{k \mathop = 0}^\infty \map {\operatorname{Res} } {-k}
}}
{{eqn | r = 2 \pi i \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k y^k} {k!}
}}
{{eqn | r = 2 \pi i e^{-y}
      | c = Power Series Expansion for Exponential Function
}}
{{end-eqn}}
We aim to show that the all but the {{RHS}} of the rectangular contour go to $0$ as we take these limits, as our result follows readily from this.
{{explain|This should be made clearer}}
The top and bottom portions of the contour can be parameterized by:
:$\map \gamma t = c \pm i R - t$
where $0 < t < c + N + \dfrac 1 2$. 
{{explain|A labelled diagram here would help}}
The modulus of the contour integral is therefore given by:
{{begin-eqn}}
{{eqn | l = \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {\map \gamma t} y^{- \map \gamma t} \map {\gamma'} t \rd t}
      | r = \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^{-\paren {c \pm i R - t} } \rd t}
}}
{{eqn | r = \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}
}}
{{eqn | r = y^{-c} \cmod {\int_0^{c + N + \frac 1 2} \map \Gamma {c \pm i R - t} y^t \rd t}
}}
{{end-eqn}}
From Bound on Complex Values of Gamma Function, we have that: 
{{begin-eqn}}
{{eqn | l = \cmod {\map \Gamma {c \pm i R - t} y^t}
      | o = \le
      | r = \frac {\cmod {c - t + i} } {\cmod {c - t + i R} } \cmod {\map \Gamma {c - t + i} y^t}
      | c = (1)
}}
{{end-eqn}}
for all $\cmod R \ge 1$. Because $\cmod R \ge 1$, we have that
{{begin-eqn}}
{{eqn | l = \frac {\cmod {c - t + i} } {\cmod {c - t + i R} }
      | o = \le
      | r = 1
}}
{{end-eqn}}
:Combining the two inequalities we obtain:
{{begin-eqn}}
{{eqn | l = \cmod {\map \Gamma {c \pm i R - t} y^t}
      | o = \le
      | r = \cmod {\map \Gamma {c - t + i} y^t}
      | c = (2)
}}
{{end-eqn}}
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. 
{{explain|I know you're referring to convergence here, but you should make it clearer.}} 
The above is enough to allow for the interchange of limits by the Dominated Convergence Theorem, thus we have:
{{begin-eqn}}
{{eqn |l = \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}
      |r = \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}
}}
{{end-eqn}}
But using Equation $(1)$ from above we see:
{{begin-eqn}}
{{eqn | l = 0 
      | o = \le
      | r = \lim_{R \mathop \to \infty} \cmod {\map \Gamma {c \pm i R - t} y^t}
}}
{{eqn | o = \le
      | r = \lim_{R \mathop \to \infty} \frac {\cmod {c - t + i} } {\cmod {c - t + i R} } \cmod {\map \Gamma {c - t + i} y^t}
}}
{{eqn | r = 0
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \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} 
      | r = \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}
}}
{{eqn | r = \lim_{N \mathop \to \infty} y^{-c} \cmod {\int_0^{c + N + \frac 1 2} 0 y^t \rd t}
}}
{{eqn | r = \lim_{N \mathop \to \infty} 0 
}}
{{eqn | r = 0
}}
{{end-eqn}} 
Thus we have that the top and bottom of the contour go to $0$ in the limit. 
The {{LHS}} 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 {{LHS}} is given as:
{{begin-eqn}}
{{eqn | l = \cmod {\int_{-R}^R \map \Gamma {\map \gamma t} y^{- \map \gamma t} \map {\gamma'} t \rd t}
      | r = \cmod {\int_{-R}^R \map \Gamma {- N - \frac 1 2 - i t} y^{- \paren {- N - \frac 1 2 - i t} } \paren {-i} \rd t}
}}
{{eqn | r = \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}
      | c = Gamma Difference Equation
}}
{{eqn | o = \le
      | r = \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
      | c = Modulus of Complex Integral
}}
{{eqn | r = \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 
}}
{{eqn | o = \le
      | r = \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
      | c = See equation (2) above
}}
{{eqn | r = \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
}}
{{eqn | o = \le
      | r = \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
}}
{{eqn | r = \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
      | c = {{Defof|Complex Modulus}}
}}
{{eqn | r = \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}
      | c = Derivative of Arctangent Function/Corollary
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \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}
}}
{{eqn | o = \le
      | r = \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}
}}
{{eqn | r = \lim_{N \mathop \to \infty} \cmod {\map \Gamma {- N + \frac 3 2} } y^{N + \frac 1 2} \frac \pi {- N + \frac 1 2}
}}
{{eqn | r = \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}
      | c = Gamma Function of Negative Half-Integer
}}
{{eqn | r = \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} 
      | c = Stirling's Formula
}}
{{eqn | r = \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} 
}}
{{eqn | r = \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} 
}}
{{eqn | r = 0
}}
{{end-eqn}}
which gives us:
{{begin-eqn}}
{{eqn | l = \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}
      | r = 0
      | c = Squeeze Theorem for Functions
}}
{{end-eqn}}
Thus we have the left, top, and bottom of the rectangular contour go to 0 in the limit, which gives us:
{{begin-eqn}}
{{eqn | l = \frac 1 {2 \pi i} \int_{c - i \infty}^{c + i \infty} \map \Gamma t y^{-t} \rd t
      | r = \frac 1 {2 \pi i} \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \oint_L \map \Gamma t y^{-t} \rd t
}}
{{eqn | r = e^{-y} 
}}
{{end-eqn}}
{{qed}}
Category:Gamma Function
Category:Complex Analysis
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