This is my last question, building off of Riemann-Von Mangoldt formula and Does $\frac{1}{\pi}\operatorname{Arg} \zeta(1/2+2\pi i t)$ have mean value $0$?. I apologize for taking a while to understand the underlying issues at hand.

Recall that $N(T)$ denotes the number of zeros of $\zeta(s)$ with imaginary part lying in $(0,T]$. For all $T > 0$ except at the points of discontinuity of $N(T)$, one has $$N(T) =1+ \frac{1}{\pi} \theta(T) + \frac{1}{\pi}\operatorname{arg}\zeta\left(\frac{1}{2}+i T\right),$$ where $\theta(T)$ is the Riemann-Siegel theta function, and where $\operatorname{arg}\zeta\left(\frac{1}{2}+i T\right)$ is chosen to vary continuously from $0$ as $s$ moves along the horizontal line from $+\infty+iT$ to $1/2+iT$. One can write

$$\frac{1}{\pi}\operatorname{arg}\zeta\left(\frac{1}{2}+i T\right) = \frac{1}{\pi}\operatorname{Arg}\zeta\left(\frac{1}{2}+i T\right)+2R(T),$$ where $R(T)$ is an integer that is $O(\log T)$, where $\operatorname{Arg}$ is the principal value of the argument, so that $$\frac{1}{\pi}\operatorname{Arg}\zeta\left(\frac{1}{2}+i T\right) = \frac{1}{\pi}\operatorname{Im}\operatorname{Log}\zeta\left(\frac{1}{2}+i T\right) \in (-1,1],$$ where $\operatorname{Log}$ is the principal branch of the complex logarithm.

I am wondering what is known about the distribution of the three functions,
$$\frac{1}{\pi}\operatorname{arg}\zeta\left(\frac{1}{2}+i T\right), \ \frac{1}{\pi}\operatorname{Arg}\zeta\left(\frac{1}{2}+i T\right), \text{ and } R(T).$$ In particular, do they each have mean $0$, and what is their variance?

Also, what is the infimum of all $T$ that is not an an imaginary part of a zero of $\zeta(s)$ such that $R(T)\neq 0$? It's kind of a "dual" to Skewes' number. EDIT: It looks like $R(T) = 1$ for $T$ the imaginary part of the $(10^{21}+7544)$-th zero of $\zeta(s)$, so this gives an upper bound for the infimum.

  • 1
    $\begingroup$ You might want to have a look at Chapter 9 of Titchmarsh's classic The theory of the Riemann zeta function. It covers a lot of discussions on the properties of the argument of $\zeta(s)$ on the critical line. $\endgroup$
    – TravorLZH
    Commented Nov 7, 2021 at 7:15

1 Answer 1


I don't know enough to answer the question for the principal branch of the argument or the question about $R(T)$, but the value distribution of $\DeclareMathOperator{\arg}{arg}\arg\zeta(1/2+it)$ is extremely well-studied in the literature.

First, following the literature in the subject, define, when $t$ is not the ordinate of a non-trivial zero, $$S(t) = \frac{1}{\pi} \arg\zeta(1/2+it) = \frac{1}{\pi}\Im\int_{\infty+it}^{1/2+it} \frac{\zeta'}{\zeta}(s) \,ds.$$ Here the integral is taken on the horizontal line from $1/2+it$ to $\infty$. When $t$ is an ordinate, the value of $S(t)$ is a matter of convention, so one takes it to be $\frac{S(t^+) + S(t^-)}{2}$.

Note that $\log\zeta$ is defined exactly analogously, except that one does not take imaginary parts and there is no normalizing factor $1/\pi$.

As you might know, the Euler product of $\zeta(s)$ is equivalent to the assertion that $$\log\zeta(s) = \sum_{n\geqslant 1} \frac{\Lambda(n)}{n^s \log n},$$ for $\Re s > 1$.

It's straightforward to see that this representation does not make sense in the critical strip $\sigma \in (0,1)$ since the right hand side doesn't converge even conditionally. Having said that, when considering statistical questions on the distribution of $\zeta$, the Euler product still exerts its influence -- see, for example Principle 1.3 in Harper's Bourbaki seminar.

In particular, using a weighted version of such a statement, Selberg showed that $$\frac{1}{2T}\int_{-T}^T S(t)^{2k}\,dt = \frac{(2k)!}{(2\pi)^{2k} k!} (\log\log T)^k (1+o(1)).$$ Further, since $S(T)$ is odd, $$\frac{1}{2T}\int_{-T}^T S(t)^{2k+1}\,dt = 0.$$ Here, $k\geqslant 0$ is an integer.

These are asymptotically precisely the moments of a normal distribution with mean $0$ and variance $\frac{1}{2\pi}\log\log T$. Since the normal distribution is characterized by its moments, this says that $S(t)$ is distributed approximately normally on $[-T,T]$. This is Selberg's central limit theorem for $S(T)$, which was first published, I believe, in the thesis of Selberg's sole PhD student, Kai Man Tsang.

In fact, Selberg's central limit theorem applies to the value distribution of $\log\zeta(1/2+it)$ on the complex plane. Informally, this states that $\log\zeta(1/2+it)$ is distributed like a standard complex Gaussian with mean $0$ and variance $\log\log T$. In other words, $\Re\log\zeta(1/2+it) = \log|\zeta(1/2+it)|$ and $\Im\log\zeta(1/2+it) = \pi S(t)$ behave like independent real Gaussians having mean $0$ and variance $\frac{1}{2}\log\log T$.

Note that since we are on the critical line, the primes are not sufficient to answer statistical questions about $\zeta$ (unlike the case for the Bohr-Jessen-esque results for fixed $\sigma > 1/2$, see Chapter 3 of Kowalski). Roughly speaking, the way one proceeds is to use a zero density estimate to control the contribution of zeros, thereby reducing the question to the statistical distribution of the Dirichlet polynomial $$\sum_{p \leqslant T} \frac{1}{p^{1/2+it}},$$ This is the harder step. The easier step is to now recall the fact that for $2\leqslant p\leqslant T$, the maps $t \mapsto p^{-it}$ behave like i.i.d. random variables uniformly taking values on the unit circle $S^1 \subset \mathbb{C}$ (this is a consequence of Kronecker-Weyl together with the fundamental theorem of arithmetic; see Principle 1.1 from Harper's Bourbaki seminar). Then, the usual central limit theorem tells you that this Dirichlet polynomial converges in distribution to a complex Gaussian with mean $0$ and variance $\sum_{p\leqslant T} \frac{1}{p} \sim \log\log T$.

Other distributional questions about $S(T)$ have also been studied. Littlewood showed that the bound $S(T) = O(\log T)$ can be improved to $$S(T) \ll \frac{\log T}{\log\log T},$$ under the assumption of the Riemann Hypothesis. Up to the quality of the implicit constant, this is still the state of the art. I believe the world record is $$S(T) \leqslant \left(\frac{1}{4}+o(1)\right) \frac{\log T}{\log\log T},$$ due to Carneiro, Chandee and Milinovich, improving the previous record of Goldston and Gonek by $1/2$. There were earlier works on this bound (Fujii, Karatsuba and Korolëv, ...) which I will leave to the mathscinet reviews to explain. The $o(1)$ is explicit, and the best bound there is due to Carneiro, Chirre and Milinovich.

A related problem is magnitude of the mean value $S_1(T)$, defined by $$S_1(T) = \int_0^T S(t) \,dt.$$ I think this was initiated by Littlewood also (op. cit.), who showed that on the Riemann Hypothesis $S_1(T) \ll \frac{\log T}{(\log\log T)^2}$. The discussion of the work of Carneiro and co-authors cited above also treats $S_1(T)$ (and more generally, an iterated integral called $S_n(T)$).

Finally, $\Omega$-results about $S(T)$, $S_1(T)$ and $S_n(T)$ are also known, and have been the subject of much recent research. The introduction of this paper of Chirre and Mahatab gives a good overview of the best known results.

Edited to add:

There's one important part of the literature that I didn't address. One important question is that of the right order of magnitude of $S(T)$. There are basically two competing conjectures.

Conjecture 1. The bound $S(T) \ll \frac{\log T}{\log\log T}$ is optimal up to the implicit constant.

Conjecture 2 (Farmer, Gonek, Hughes). One has, $$\limsup_{T\to\infty} \frac{S(T)}{\sqrt{\log T\log\log T}} = \frac{1}{\pi \sqrt{2}}.$$

I think some people believed Conjecture 1 for a long time mostly because, as I indicated above, that bound has not been improved under RH. Most people in the area seem to believe Conjecture 2 now, though (Farmer-Gonek-Hughes make a very compelling case via several heuristic arguments).

  • $\begingroup$ Many thanks for all of the details and references! I am still waiting on a couple of them from my library. I'm surprised that no one has seemed to study the integer-valued function $R(T)$, which seems very likely to have mean $0$ and captures the size of $S(T)$. It seems to be like an dual analogue of the Mertens function. Do you have any idea what is the infimum of $T$ such that $R(T) \neq 0$? $\endgroup$ Commented Nov 9, 2021 at 11:33
  • $\begingroup$ Sorry, I have no ideas about $R(T)$. I tried seeing if I could unpack Lucia's comment about the sign of the Hardy $Z$-function in your earlier question, and it certainly seems related, but I couldn't draw out the explicit connection. By the way, I have e-copies of almost everything I mentioned in the answer, so feel free to reach out by email and I can share whichever papers you want. $\endgroup$ Commented Nov 9, 2021 at 13:39

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