# What's the easiest way to prove the Riemann Zeta function has any zeros at all on the critical line?

I learned in my intro to complex analysis class that the Riemann Zeta has trivial zeros at the negative even integers; this follows from the Riemann Zeta Reflection identity and the behavior of the $$\Gamma$$ function on negative integers. According to Wikipedia, Hardy proved there are infinitely many zeros on the critical line $$Re(z) = \frac{1}{2}$$, and apparently Riemann "checked that a few of the zeros lay on the critical line..." But to bridge the complexity gap here, is there an easy way to see 1. that $$\zeta$$ has other zeros at all besides these trivial ones, and 2. that it has any zeros at all on the critical line? How can I verify at home a prospective root?

The easiest way is to refer to the Hardy $$Z$$ function, $$Z(t) = e^{i \theta(t)} \zeta(1/2 + it),$$ where $$e^{i \theta(t)} = \pi^{-i t / 2} \frac{\Gamma(1/4 + it/2)}{\lvert \Gamma(1/4 + it/2) \rvert}.$$ It's not hard to show that $$Z(t)$$ is real when $$t$$ is real, and clearly $$\lvert Z(t) \rvert = \lvert \zeta(1/2 + it) \rvert$$. Hence they have the same zeros.
Thus one way to show that $$\zeta(s)$$ has a zero on the half-line is to find a value of $$Z(t)$$ that is positive, and another value that is negative: then there is a zero somewhere in the middle. Further, repeatedly computing and bisecting allows you to compute a zero to any desired precision.
• @quarague Do we take for granted that we can compute $\Gamma(s)$? I assume so. If not, then good approximations to the gamma function can be approximated reasonably well via numerical integration. A better approximation comes from the Lanczos approximation. For $\zeta(1/2 + it)$, using the simple eta function $\eta(s) = \sum_{n \geq 1} (-1)^{n+1}/n^s$ and the immediate-upon-consideration relationship $\eta(s) = (1 - 2^{1-s}) \zeta(s)$ lets you compute $\zeta(1/2 + it)$, as $\eta(s)$ converges for $\mathrm{Re}(s) > 0$. Euler acceleration improves this. Euler-Maclaurin... (continued) Commented Jul 16 at 19:46
• (continued)... Euler-Maclaurin would also work. There are a wealth of methods to compute $\zeta$. The approximate functional equation works well, and for $\zeta(s)$ there is a particularly nice form called the Riemann-Siegel formula. This is less obvious, but numerically much better behaved. In practice, one would use the Odlyzko-Schonhage algorithm to find lots of zeros to high precision quickly. Commented Jul 16 at 19:48