Is the functional equation for the inverse of Riemann zeta function valid for any point in the critical strip?

Question

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?

Answer 1

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.