Is there any sort of (closed form preferably, though if not, it's fine) function for $\zeta(\frac12+it)$ where $\zeta$ is the Riemann zeta function? Anything is welcome, so I can take it from there. Upper bounds, lower bounds and asymptotics are also accepted.

$\begingroup$ How about the Riemann Siegel formula? $\endgroup$– user156885May 5, 2020 at 6:50

$\begingroup$ @zz7948 I'm not sure how that helps. $\endgroup$– DUO LabsMay 5, 2020 at 16:19

$\begingroup$ @Dave related to approximating zeta. Good for numerical approximations and maybe the best we have to a ’closed form’ in some situations. Taking it with the ’anything is welcome’ attitude $\endgroup$– user156885May 5, 2020 at 16:54

$\begingroup$ @zz7948 Yeah, but that was for zeta, not for the absolute value of it. $\endgroup$– DUO LabsMay 5, 2020 at 17:23
1 Answer
It is not known whether, for $t>0$ large, $\zeta(1/2+it)$ is smaller than $t^{1/7}$, while it is conjectured that $1/7$ can be replaced by any positive exponent. This should answer your question. For more information, consult Titchmarsh's book on the Riemann zeta function, there is a whole chapter on this topic. Another good starting point is the relevant Wikipedia entry.


$\begingroup$ @QuoteDave: There are hundreds of papers on the subject. As a starter, read Titchmarsh's chapter and the Wikipedia entry (in my post). See also the recent work of Soundararajan, Harper etc. on the arXiv. This is one the hot topics of analytic number theory. $\endgroup$ May 5, 2020 at 3:38

$\begingroup$ Though when they say "$c_{k,j}$ as some constant", what kind of constants are they talking about? $\endgroup$– DUO LabsMay 5, 2020 at 3:41

2$\begingroup$ @QuoteDave: That asymptotic formula is only conjectured. As the Wikipedia page says, a conjectured formula for $c_{k,j}$ was given by Keating & Snaith (2000). The best unconditional upper bounds appear in the table in the Wikipedia page. Please study the literature. It is vast and easy to find. $\endgroup$ May 5, 2020 at 3:46

$\begingroup$ No, I am really only interested in $k=1$, which has been proven, and the Wikipedia page never gave any values of $c$. $\endgroup$– DUO LabsMay 5, 2020 at 3:49