TITLE: Why does zeta have infinitely many zeros in the critical strip?
QUESTION [16 upvotes]: I want a simple proof that $\zeta$ has infinitely many zeros in the critical strip.
The function $$\xi(s) = \frac{1}{2} s (s-1) \pi^{\tfrac{s}{2}} \Gamma(\tfrac{s}{2})\zeta(s)$$ has exactly the non-trivial zeros of $\zeta$ as its zeros ($\Gamma$ cancels all the trivial ones out). It also satisfies the functional equation $\xi(s) = \xi(1-s)$.
If we assume it has finitely many zeros, what analysis could get a contradiction?
I found an outline for a way to do it here but I can't do the details myself: https://mathoverflow.net/questions/13647/why-does-the-riemann-zeta-function-have-non-trivial-zeros/13762#13762

REPLY [2 votes]: (I think this is an elementary proof)
everybody here knows that the zeros of $\zeta(s)$ are useful mainly for the Riemann explicit formula :
$$\psi(x) = \sum_{p^k \le x} \ln p = x - \sum_{\rho} \frac{x^\rho}{\rho}-\ln(2 \pi)- \frac{1}{2}\ln(1-x^{-2})$$
suppose there is a finite number of non-trivial zeros, then $\frac{d}{dx}\psi(x)$ could not be the distribution : $$\sum_{p \in \mathcal{P}} \sum_{k=1}^\infty \ \delta(x-p^k) \ \ln p $$