Poles of the gamma function in the zeta functional equation

Question

I doubt it is rigorous to derive $$\zeta (s)=2^s\pi ^{s-1}\sin \frac{\pi s}{2} \, \Gamma (1-s)\zeta(1-s)$$ from $$\pi ^{-\frac{s}{2}}\Gamma \left(\frac{s}{2}\right) \zeta (s)=\pi ^{\frac{s-1}{2}}\Gamma \left(\frac{1-s}{2}\right)\zeta (1-s)$$ using Euler's reflection formula, since $$\Gamma (s)\Gamma (1-s)=\frac{\pi}{\sin \pi s}$$ is correct only for non-integer $s$, so how can we assume all complex $s$ in the functional equation (the one involving $\sin$)?

Also, if $$t(s)=\sin \frac{\pi s}{2} \, \Gamma (1-s),$$ then $t(2)$ is undefined, only $\lim_{s\to 2}t(s)$ exists. Doesn't this contradict the fact that $\zeta (2)$ exists?

It seems I'm missing something very important and would appreciate any help.

Answer 1

In complex analysis :

If $f$ is analytic (around $s=a$) with a simple zero at $s=a$ and $g$ is meromorphic (around $s=a$) with a simple pole at $s=a$ then $fg$ is analytic and non-zero at $s=a$.

You can call $s=a$ a removable singularity of $fg$.

Note I didn't write $f(s)g(s)$ (the quotient of two complex numbers) I wrote $fg$ (the product of two meromorphic functions).