Explicit example of a non-trivial zero of Riemann zeta function

Question

Assume we are given the Riemann zeta function on $\mathrm{Re}(s) > 0$ by:

$$\zeta(s) = \dfrac{s}{s-1} - s\int_1^{\infty} \dfrac{\{u\}}{u^{s+1}}du$$

My question is: can you give me explicitely a real number $t>0$ such that $$\zeta(1/2 + it) = 0$$ (and providing a proof that this is exactly a zero of $\zeta$).

I saw questions like Show how to calculate the Riemann zeta function for the first non-trivial zero or Proving a known zero of the Riemann Zeta has real part exactly 1/2, but none of them seem to give a concrete and exact example (I don't want to have approximations, nor to use a computer).

It is actually possible to have an exact value for (at least) one zero of $\zeta$ ? Maybe this is not possible, this is why I'm asking.

Answer 1

For $Re(s) > 1$ let $$\xi(s) = 2\pi^{-s/2} \Gamma(s/2) \zeta(s)=\int_0^\infty x^{s/2-1} (\theta(x)-1)dx, \qquad \theta(x) = \sum_{n=-\infty}^\infty e^{-\pi n^2 x}$$ With the Poisson summation formula we find that $\theta(1/x) = x^{1/2}\theta(x)$ and $$\xi(s) = \int_0^1+\int_1^\infty x^{s/2-1} (\theta(x)-1)dx$$ $$= \frac{1}{s-1}-\frac{1}{s}+\int_1^\infty (x^{s/2-1}+x^{(1-s)/2-1}) (\theta(x)-1)dx = \xi(1-s)$$ which is true for any $s$. Also $\xi(\overline{s}) = \overline{\xi(s)}$ so that

$Z(t) = \xi(1/2+it)$ is a function $\mathbb{R} \to \mathbb{R}$. It has a zero at every sign change. The Riemann hypothesis is that it doesn't have any other zero. Its 1st sign change is at $t \approx 14.134725$

Answer 2

Set $s=14.000000000000000000000000000000...i$ with 1000 zeros after the decimal point. Set $n=21$ and set $c = 1 + 1/10^{40}$; With those parameters compute this formula:

$$s-\frac{n \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}$$

What you will get is the 25 first decimal digits of the first Riemann zeta zero:

0.50000000000000000000000055508907479219367612957050478295942858083862
3727033228398609021142110650620136997773667771872221905026127340639625
41218507480832131294005829437
+
14.134725141734693790457251915896759601972505820234600660252328557362
5629956990194271674005286735176937891872097245657731536209606798029380
8035224527780328742481096881866 I

Of course ideally: $n \rightarrow \infty$ and $c \rightarrow 1$

https://mathoverflow.net/q/368533/25104
Show that when applying these substitution rules the result is an alternating sum with binomial coefficients in the numerators.

(*Mathematica*)
(*start*)
Clear[n, k, s, c];
n = 21;
s = N[14*I, 1000];
c = 1 + 1/10^40;
s - n*(1/Zeta[c]*
    Sum[(-1)^(k - 1)*
       Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n}]/
     Sum[(-1)^(k - 1)*
       Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}])
(*end*)