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:  {{begin-eqn}} {{eqn	| l = \lim_{N \to \infty} \lim_{R \to \infty} \oint_L \Gamma \left({t}\right) y^{-t} \, \mathrm d t 	| r = 2 \pi i\displaystyle \sum_{k\mathop = 0}^{\infty} \operatorname{Res} \left({-k}\right) }} {{eqn	| r = 2 \pi i \sum_{k\mathop = 0}^{\infty} \frac{ \left({-1}\right)^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 righthand side 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: :$\gamma\left({t}\right)= c\pm iR - t$ where $0 < t < c + N + \frac 1 2$.  {{explain|A labelled diagram here would help}} The modulus of the contour integral is therefore given by: {{begin-eqn}} {{eqn	| l = \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 	| r = \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 }} {{eqn	| r = \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 }} {{eqn	| r = y^{-c} \left\vert \int_0^{c + N + \frac 1 2} \Gamma \left({c\pm iR - t}\right) y^t \rd t \right\vert }} {{end-eqn}} From Bound on Complex Values of Gamma Function, we have that:  {{begin-eqn}} {{eqn |l = \left\vert \Gamma \left({c\pm iR - t}\right) y^t \right\vert       |o = \leq       |r=  \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       |c= (1) }} {{end-eqn}} for all $|R|\geq 1$. Because $|R|\geq 1$, we have that {{begin-eqn}} {{eqn |l = \frac{\left\vert c-t+i\right\vert}{\left\vert c-t+iR \right\vert}       |o = \leq       |r = 1 }} {{end-eqn}} :Combining the two inequalities we obtain: {{begin-eqn}} {{eqn |l = \left\vert \Gamma \left({c\pm iR - t}\right) y^t \right\vert       |o = \leq       |r =  \left\vert \Gamma \left({c-t+i}\right) y^t \right\vert       |c = (2) }} {{end-eqn}} 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.  {{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 \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        |r = \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 }} {{end-eqn}} But using Equation $(1)$ from above we see: {{begin-eqn}} {{eqn | l=0        | o= \leq        | r= \lim_{R \to \infty}  \left\vert \Gamma \left({c\pm iR - t}\right) y^t \right\vert }} {{eqn |o= \leq       |r= \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 }} {{eqn |r=0 }} {{end-eqn}} 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: {{begin-eqn}} {{eqn |l = \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        |r = \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 }} {{eqn |r= \lim_{N \to \infty} y^{-c} \left\vert \displaystyle \int_0^{c+N+\frac{1}{2} } 0 y^t \rd t \right\vert }} {{eqn |r=  \lim_{N \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: :$\gamma\left({t}\right) = N - \dfrac 1 2 - it$ where $t$ runs from $-R$ to $R$.  Thus the absolute value of integral of the left-hand side is given as: {{begin-eqn}} {{eqn | l = \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       | r = \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 }} {{eqn | r = \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       | c = Gamma Difference Equation }} {{eqn | o = \leq       | r = \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        | c = Modulus of Complex Integral }} {{eqn | r = \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  }} {{eqn | o = \leq       | r = \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       | c = See equation (2) above }} {{eqn | r = \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 }} {{eqn | o = \leq       | r = \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 }} {{eqn | r = \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       | c = Definition of Complex Modulus }} {{eqn | r = \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}       | c = Derivative of Arctangent Function/Corollary }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn |l = 0       |o = \leq       |r = \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 }} {{eqn |o=\leq       |r= \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} }} {{eqn| r= \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} }} {{eqn| r= \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}      | c= Gamma Function of Negative Half-Integer }} {{eqn|r= \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}       |c= Stirling's Formula }} {{eqn|r= \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}  }} {{eqn|r= \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}  }} {{eqn|r= 0 }} {{end-eqn}} Which gives us: {{begin-eqn}} {{eqn| l= \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      | r= 0      | c= Squeeze Theorem }} {{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= \dfrac{1}{2\pi i} \displaystyle \int_{c-i\infty}^{c+i\infty} \Gamma \left({t}\right) y^{-t} \rd t      |r=  \dfrac{1}{2\pi i} \lim_{N \to \infty} \lim_{R \to \infty} \oint_L \Gamma \left({t}\right) y^{-t} \, \mathrm d t }} {{eqn| r = e^{-y}  }} {{end-eqn}} {{qed}} Category:Gamma Function Category:Complex Analysis 3g44dfic49983k70ynlfr685wrahckp