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.
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$\begingroup$ How about the Riemann Siegel formula? $\endgroup$– user156885Commented May 5, 2020 at 6:50
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$\begingroup$ @zz7948 I'm not sure how that helps. $\endgroup$– DUO LabsCommented May 5, 2020 at 16:19
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$\begingroup$ @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 $\endgroup$– user156885Commented May 5, 2020 at 16:54
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$\begingroup$ @zz7948 Yeah, but that was for zeta, not for the absolute value of it. $\endgroup$– DUO LabsCommented May 5, 2020 at 17:23
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1 Answer
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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.
answered May 5, 2020 at 3:34
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$\begingroup$ Are there any unconditional results? $\endgroup$– DUO LabsCommented May 5, 2020 at 3:35
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$\begingroup$ @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. $\endgroup$ Commented May 5, 2020 at 3:38
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$\begingroup$ Though when they say "$c_{k,j}$ as some constant", what kind of constants are they talking about? $\endgroup$– DUO LabsCommented May 5, 2020 at 3:41
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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. $\endgroup$ Commented May 5, 2020 at 3:46
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$\begingroup$ 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 LabsCommented May 5, 2020 at 3:49