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Let $\epsilon > 0$, and $ n \in \mathbb{Z}^{+} $. Let $C_{n}$ be a positively oriented polygonal line that is from $-n + 1/2 - i \epsilon$ to $ 1/2 - i \epsilon$ and from $ 1/2 - i \epsilon$ to $ 1/2 + i \epsilon$ and from $ 1/2 + i \epsilon$ to $-n + 1/2 + i \epsilon$. And now define polygonal curve $C_{R}$ in the same way above but replacing $n$ with a real number $R>0$. Let $\Gamma$ be Gamma function. Then is it $$ \int_{C_{n}} \Gamma = \int_{C_{R}} \Gamma $$ as $n \to \infty$ ?

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If $\lim_{R\to\infty}f(R)$ exists, then $\lim_{R\to\infty}f(R)=\lim_{n\to\infty}f(n)$ (note that it isn't even easy to make the difference between both cases notationally clear, but you know what is meant).

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The answer by Hagen von Eitzen has an important clause: if $\lim_{R\to\infty} f(R)$ exists. One must worry at least a bit about passing by negative integers, where $\Gamma$ has poles. Lengthening the (unclosed) contour of integration near a pole has a potential to perturb the integral by a non-negligible amount. But looking at this graph of $|\Gamma|$, one sees that the spikes get rather thin as we move in the negative direction, and therefore $\Gamma(x\pm i\epsilon)$ should be a nice function for any fixed $\epsilon$, when $x$ goes far enough in the negative territory.

From Wikipedia article on Gamma

The functional equation $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}$$ confirms that the plot from Wikipedia is not lying. The right hand side is bounded on $\operatorname{Im} z =\pm \epsilon$ by $\pi/\sinh \epsilon$, and $|\Gamma(1-(x\pm i\epsilon))|$ grows superexponentially as $x\to -\infty$ (e.g., by repeated application of $\Gamma(z+1)=z\Gamma(z)$).

Conclusion: the limit $\lim_{R\to\infty} \int_{C_{R}} \Gamma$ exists.

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