Sequential Limit of Line Integral Is The Same As The Usual Limit of Line Integral? (Gamma Function Related)

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 $n \to \infty$ ?

-

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).
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.
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.