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It is known that in the critical strip $s\in \{0<\mathrm{Re}(s)<1\}$,Riemann zeta function satisfies the following functional equation:

$$\zeta(s)=\chi(s)\zeta(1-s),\tag{1}$$ $$\chi(s)=\frac{\pi^{-s/2}\Gamma(s/2)}{\pi^{-(1-s)/2}\Gamma((1-s)/2)},\tag{2}$$

Because $0<|\chi(s)|<\infty$, when $\zeta(s)=0$, we have $\zeta(1-s)=0$. So (1) becomes $0=0$ which is still valid.

When $\zeta(s)\not=0\not=\zeta(1-s)$, we can rewrite (1) as:

$$\frac{1}{\zeta(1-s)}=\chi(s)\frac{1}{\zeta(s)},\tag{3}$$

Because $0<|\chi(s)|<\infty$, when $\zeta(s)=0$, we have $\zeta(1-s)=0$. So (3) becomes $\infty=\infty$.

Question is $\infty=\infty$ considered a valid identity? What kind of limiting process is necessary to make it valid?

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    $\begingroup$ the keyword is equality of en.wikipedia.org/wiki/Meromorphic_function . The en.wikipedia.org/wiki/Analytic_continuation theorem is that if $f,g$ are meromorphic and $f(z) = g(z)$ on an open $U \subset \mathbb{C}$ or even a line segment, then $f = g$ on the whole complex plane (in the sense of memorphic functions, i.e. everywhere except at the poles) $\endgroup$
    – reuns
    Jul 4, 2016 at 18:08
  • $\begingroup$ @user1952009: Thanks for the explanation. $\endgroup$
    – mike
    Jul 4, 2016 at 19:04

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Your question has little to do with the $\zeta$ function. Here is a rewriting of your question in simpler terms:

"We have the identity $x=x$ so if $x\not=0$ then we can write $\frac{1}{x} = \frac{1}{x}$. Now if I take $x=0$ then we get $\infty=\infty$. Is this a valid identity?"

When we say that $f(x) = g(x)$ is an identity we mean that the equality holds for all $x$ in the domain of $f,g$ (or in some specified domain). When you divide by $x$ above you are (as you write) considering $x\not=0$ so taking $x=0$ does not make sense as $x=0$ is not in the domain of this latter equation. The case "$\infty=\infty$" is never an issue here.

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