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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.

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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.

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    $\begingroup$ 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. $\endgroup$ – José Carlos Santos Oct 19 '19 at 20:46
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    $\begingroup$ @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... $\endgroup$ – Sangchul Lee Oct 19 '19 at 20:46

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