Why the Riemann Zeta function has nontrivial zeros

I know that the Riemann Zeta function has an infinite number of zeros on the critical line $$\sigma = 1/2$$;

that it is possible to determine how many zeros the Riemann Zeta function has on any interval $$[T, T+H]$$;

that the Riemann Zeta function is defined:

by improper integral $$\zeta(s) = s\int_{0}^{\infty}{\frac{[x]-x+\frac{1}{2}}{x^{s+1}}}dx$$;

or by other improper integral $$\Gamma(s)\zeta(s) = \int_{0}^{\infty}{\frac{x^{s-1}dx}{e^x-1}}$$

has the first approximate equation $$\zeta(s)=\sum_{n\le x}{\frac{1}{n^s}}-\frac{x^{1-s}}{1-s}+\mathcal{O}(x^{-\sigma})$$;

and the second approximate equation $$\zeta(s)=\sum_{n\le x}{\frac{1}{n^s}}+\chi(s)\sum_{n\le y}{\frac{1}{n^{1-s}}}+\mathcal{O}(x^{-\sigma})+\mathcal{O}(|t|^{1/2-\sigma}y^{\sigma-1})$$;

I know how to calculate the non-trivial zeros of the Riemann Zeta function...

But I can't understand why the Riemann Zeta function has nontrivial zeros and why they lie on the critical line $$\sigma = 1/2$$?

• Check this for various approaches and answers- mathoverflow.net/questions/13647/… – MathematicianByMistake May 8 at 16:06
• Let $\xi(s) = s(1-s)\pi^{-s/2} \Gamma(s/2) \zeta(s)$ then $\xi(s) = \xi(1-s)$ and $\xi(s) = \overline{\xi(\overline{s})}$ and it is entire and its zeros are the non-trivial zeros of $\zeta(s)$. So $\xi(1/2+it)$ is real and it has a zero at every sign change. In his book Titchmarsh shows a simple contradiction when assuming $\xi(1/2+it)$ has the same sign for $|t|$ large enough, thus there are infinitely many non-trivial zeros on $\Re(s)=1/2$. – reuns May 8 at 16:07
• Also $\xi$ has growth order $e^{|s|^{1/2+\epsilon}}$, so it is not of the form $P(s) e^{f(s)}$ with $P$ a polynomial and $f$ entire, thus it has infinitely many zeros. A little more work shows the increase of $arg \log \zeta(s), s \in [1/2+it, 2+it]$ is $O(\log t)$ so the density of zeros is $arg \Gamma((1/2+iT)/4)+O(\log T) \sim C T \log T$. The next steps are harder (density of zeros on $\Re(s) \ge \sigma > 1/2$, positive proportion of zeros on $\Re(s) = 1/2$). – reuns May 8 at 16:07
• I have a beautiful proof that the non-trivial zeros lie on the line $\sigma=\frac12$ but it's too long to fit in a comment. – saulspatz May 8 at 16:30