# Riemann-Von Mangoldt formula, revised question

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.

• 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. Commented Nov 7, 2021 at 7:15

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.

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, $$\limsup_{T\to\infty} \frac{S(T)}{\sqrt{\log T\log\log T}} = \frac{1}{\pi \sqrt{2}}.$$
• 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 $R(T) \neq 0$? Commented Nov 9, 2021 at 11:33
• 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. Commented Nov 9, 2021 at 13:39