All the sources I have read talk about continuation from $Re(s)>1$ to $Re(s)>0$ then $Re(s)<0$ $(s\neq 1)$. What about $Re(s)=0$? Where does that go?
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1The functional equation implies a symmetry between the cases $\Re(s)\ge 1$ and $\Re(s)\le 0$. For some results see for example $(9)$ here. – Raymond Manzoni Apr 27 '16 at 22:57
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the functional equation is proven first for $Re(s) \in \ ]0,1[$, but then by analytic continuation it is valid everywhere (except at $s=1$ or $s=0$ of course).
you can also analytically continuate $\zeta(s)$ from $$\Gamma(s) \zeta(s) = \int_0^1 \left(\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!} x^k\right)x^{s-2}dx + \int_1^\infty \frac{x^{s-1}}{e^x-1}dx + \sum_{k=0}^K \frac{B_k}{k!} \frac{1}{s+k-1}$$
valid for every $Re(s) > - K$ and not a negative integer (with $B_k$ the Bernouilli numbers)
and don't forget that in complex analysis, we don't consider functions but analytic continuation of functions.
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