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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.

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    $\begingroup$ Why do you expect the closed form of $t$ to exist or be known? $\endgroup$ Mar 12 '17 at 13:30
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    $\begingroup$ I'm not necessarily asking about a closed form (even if this would be better). I just want that you define me precisely the $t$ (the definition can involve limits, integrals, ... if needed), and then I would like to have a proof of $\zeta(1/2 + it)=0$. $\endgroup$
    – Alphonse
    Mar 12 '17 at 13:31
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    $\begingroup$ Riemann-Siegal formula is most likely how Riemann found the first couple zeros pre-computers [[source]](mathoverflow.net/questions/142548/…) and Euler Maclaurin summations are often used [[source]](math.dartmouth.edu/archive/m56s13/public_html/Nguyen_proj.pdf). Both use approximations, but they are good approximations and give an error bound. $\endgroup$
    – Dando18
    Mar 12 '17 at 13:41
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    $\begingroup$ It's pretty hard to get any kind of exact expression for a random real or complex number, basically because there are uncountably many of them and we can only build countably many exact expressions. It would be quite the miracle if there were exact expressions for the Riemann zeros. (By the way, you seem to be asking if the zeros of the Riemann zeta function are periods: maths.ed.ac.uk/~aar/papers/kontzagi.pdf) $\endgroup$ Mar 12 '17 at 15:03
  • $\begingroup$ I found this : math.stackexchange.com/questions/190080/… $\endgroup$
    – Alphonse
    May 24 '17 at 18:35
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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 $t \approx 14.134725$

enter image description here

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    $\begingroup$ the plotted function is $Z(t) = \frac{\xi(1/2+it)}{\pi^{-1/2} |\Gamma((1/2+it)/2)|}$ but it doesn't change anything $\endgroup$
    – reuns
    Mar 13 '17 at 12:28
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    $\begingroup$ Thank you for your answer! However you don't give an exact value of $t$, only an approximation. That's still interesting. $\endgroup$
    – Alphonse
    Mar 13 '17 at 17:46
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    $\begingroup$ @Alphonse You can't do better. It is believed $t_1$ is "a new mathematical constant", as $\pi,e,\gamma$.. $\endgroup$
    – reuns
    Mar 13 '17 at 18:10
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    $\begingroup$ @Alphonse, what do you mean by "exact value"? The imaginary part of the first zero, $14.134725\ldots$, is conjectured to be a transcendental number, and it seems unlikely that there is a "nice" closed-form expression for it. $\endgroup$ Mar 13 '17 at 20:34
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    $\begingroup$ I think it's a perfectly fine answer to a poorly posed question. $\endgroup$ Mar 13 '17 at 20:54
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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*)
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