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I want to evaluate the following limit: $$\lim_{s \to 1}\; \Gamma \left( \frac{1-s}{2} \right) (s-1).$$ I know that the gamma function has simple poles at $-n$ for $n \in \mathbb{N}_0$ with residue $\frac{(-1)^n}{n!}$. Also, I know that if a function $f$ has a simple pole at $a$, then $\operatorname{Res}_a f= \lim_{z \to a} (z-a) f(z)$. If $\Gamma(z)$ has a simple pole at $z=0$, then $\Gamma (\frac{1-s}{2})$ has a pole at $s=1$. If it was a simple pole, the limit would be $-1$, however, wolframalpha tells me the limit must be $-2$ and in the context I am using this limit, $-2$ must be the correct result.

Where did I make a mistake? Thanks for any help in advance.

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up vote 13 down vote accepted

$$\lim_{s \to 1}\; \Gamma \left( \frac{1-s}{2} \right) (s-1)=-2\cdot \lim_{s \to 1}\; \Gamma \left( \frac{1-s}{2} \right) \frac{1-s}{2}=$$ $$=-2\cdot\lim_{x\to 0}\ \Gamma(x)\ x=-2\cdot\lim_{x\to 0}\Gamma(x+1)=-2\cdot \Gamma(1)=-2$$

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+1 Very nice and smooth. –  DonAntonio Jan 14 '13 at 13:34
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The existence of a pole of $\Gamma$ at $0$ can be translated as the fact that $z\Gamma(z)\to1$ when $z\to0$. Hence $\frac{1-s}2\Gamma\left(\frac{1-s}2\right)\to1$ when $s\to1$, that is, $(s-1)\Gamma\left(\frac{1-s}2\right)\to-2$.

Your mistake could be to confuse $\frac{1-s}2\Gamma\left(\frac{1-s}2\right)$ and $(s-1)\Gamma\left(\frac{1-s}2\right)$.

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