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

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.

Solutions Collecting From Web of "Explicit example of a non-trivial zero of Riemann zeta function"

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 $t \approx 14.134725$

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