# Analytic continuation in a proof in Apostol's analytic number theory textbook

In Apostol's Introduction to Analytic Number Theory, the following equation is derived for real $$s>1$$: $$\Gamma(s)\zeta(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,\mathrm{d}x$$ Then, the fact that both sides are analytic on the half-plane with $$\Re(s)>1$$ is used to extend this result to all complex $$s$$ with $$\Re(s)>1$$. All it says in the textbook is "by analytic continuation" but it doesn't actually eplain what allows this result to be extended. I know what analytic continuation is, and I know about the principle of analytic continuation, but the real line is not open in the complex plane, so surely this doesn't work; just because two analytic functions are equal for real $$s>1$$ doesn't mean that they are equal on the half-plane with $$\Re(s)>1$$ even if they are both defined there. Can somebody please explain what allows this result to be extended?

EDIT: I thought that the identity theorem (what I called the principle of analytic continuation) only works when the two functions are equal on an open subset of the complex plane, but, as a couple of people have helpfully pointed out, it applies when the two are equal on any set containing at least one non-isolated point. Clearly, no point of $$(1,\infty)$$ is isolated, so it can be applied here.

Since the equality$$\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^{s-1}}{e^x-1}\,\mathrm ds$$holds when $$s\in(1,\infty)$$, and since $$s\mapsto\Gamma(s)\zeta(s)$$ and $$s\mapsto\int_0^\infty\frac{x^{s-1}}{e^x-1}\,\mathrm ds$$ are analytic functions defined on $$\{z\in\mathbb C\mid\operatorname{Re}z>1\}$$, then we have$$\operatorname{Re}z>1\implies\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^{s-1}}{e^x-1}\,\mathrm ds$$by the identity theorem.
• Wrong! It applies when the two are equal on a set with at least one non-isolated point. And no point of $(1,\infty)$ is isolated. – José Carlos Santos Oct 19 '19 at 20:46
• @windingnumberone, $\Omega=\{z\in\mathbb{C}:\operatorname{Re}(z)>1\}$ is an open subset of $\mathbb{C}$ and $(1,\infty)$ is a subset of $\Omega$ having an accumulation point in $\Omega$, so the identity theorem works... – Sangchul Lee Oct 19 '19 at 20:46