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This is a subtle point about analytic continuation. Let $\Gamma(s)$ be the analytic continuation of $\gamma(s) := \int_0^\infty e^{-t}t^{s-1}dt$ to $\Bbb C \setminus \Bbb Z_{<0}$, the latter function being only defined for $\operatorname {Re}(s)\ge 0$. Notice that one can prove from the definition of $\gamma(s)$ that it satifies the functional equation $\gamma(s+1) = s\gamma(s)$ where it is defined.

Stein and Shakarchi claim on page 162 of their book complex analysis that because $\Gamma (s)$ and $\gamma(s)$ agree on $\operatorname {Re}(s)\ge 0$ and $\Gamma(s)$ is holomorphic away from its poles, the functional equation stated above is also true $\Gamma(s)$ wherever it is holomorphic, and takes the form $\operatorname{res}_{s=-n} \Gamma(s+1)=(-n)\operatorname{res}_{s=-n} \Gamma(s)$ at the poles of $\Gamma(s)$.

My question is whether the above reasoning is really correct, and, if so, why. I see why it is true for the gamma function because $\Gamma(s)$ was extended from $\gamma(s)$ using precisely the functional equation of $\gamma(s)$, but could one know that this is true if one extended $\gamma(s)$ by a power series? In other words, (why) is it in general true that any functional equation $E(f(s),s)=0$ of a holomorphic function $f(s)$ defined on a connected domain $\omega$ will also hold for the analytic continuation $F(z)$ of $f(s)$ into a larger connected domain $\Omega \supset \omega$.

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I'm not certain if this is the difficult point for you.

Since the holomorphic function $h(s) = \Gamma(s+1) - s\Gamma(s)$ satisfies $h(s) = \gamma(s+1) - s\gamma(s) = 0$ for $\textrm{Re}(s) > 0$, we have $h(s) = 0$ throughout its domain, due to the following principle of analytic continuation.

If a holomorphic function $h$ is defined on a connected open set $\Omega$, and $h(z) = 0$ for all $z$ belonging to some subset of $\Omega$ having a point of accumulation within $\Omega$, then $h$ is identically zero throughout $\Omega$.

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  • $\begingroup$ Thank you for your answer. You are right, this shouldn't have been difficult; I simply had this gap from the first time I studied complex analysis a couple years ago, and, for some reason, I wasn't able to fix it alone later. Well, it was partly because I heard some professors say that you always have to check that the extended function also satisfies the functional equation. I don't know what they meant! Maybe they didn't really understand my question. $\endgroup$ – Rodrigo Jan 14 '14 at 13:23

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