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Define $N_\Delta(T)$ to be the number of sign changes of $\psi(x) - x$ in the interval $[1, T]$.

Landau's Theorem says $N_\Delta(T)$ is $\Omega(\log T)$ [1].

But perhaps that estimate is too crude. Is the main term of $N_\Delta(T)$ known? Or are only strict upper and lower bounds known?

What type of machinery is used to determine something of this nature?


[1] Emil Grosswald "Oscillation Theorems of Arithmetical Functions" Transactions of the American Mathematical Society 126 (1967) pp. 7.

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An asymptotic for the number of sign changes is not known, and indeed only a lower bound of $c \log T$ is known. There have been small improvements in the constant $c$ that is allowed here (see Kaczorowski). For recent work related to this, see Montgomery and Vorhauer, who show that the remainder term changes sign in any interval $[x,2.02x]$ once $x$ is large enough. They also speculate that the right order for the number of sign changes up to $T$ is $\sqrt{T}$ (every once in a while $\psi(x)-x$ will be very small, and near that point there'll be many sign changes). One more recent reference is a paper of Kaczorowski and Wiertelak.

Also one shouldn't really expect an asymptotic formula for the number of sign changes -- if $\psi(x)-x$ is large and positive (on scale $\sqrt{x}$) then it is likely to stay positive at $1.01x$ as well. Montgomery and Vorhauer speculate that
$$ 0< \liminf_{T\to \infty} \frac{N_{\Delta}(T)}{\sqrt{T}} < \limsup_{T\to\infty} \frac{N_{\Delta}(T)}{\sqrt{T}} < \infty. $$

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  • $\begingroup$ Great, thanks! Do you have a source for knowing the main term is unknown, or is that something you "just know"? $\endgroup$ May 30, 2016 at 23:06
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    $\begingroup$ Well the paper by Montgomery and Vorhauer clearly states this, and this hasn't changed in the last ten years. $\endgroup$
    – Lucia
    May 30, 2016 at 23:07
  • $\begingroup$ Ah, I missed that part. Thanks again. $\endgroup$ May 30, 2016 at 23:09
  • $\begingroup$ Without assuming the RH it seems obvious we can't estimate the number of sign changes of $\displaystyle \frac{\psi(e^u)-e^u}{e^{u/2}}=\mathcal{O}(e^{-u/2})+\sum_\rho \frac{e^{(\rho-1/2) u}}{\rho}$. But assuming the RH, the number of sign changes of $\displaystyle Re(\sum_{n \ge 1} \frac{e^{i \frac{2 \pi n}{\ln n} u}}{1/2+i \frac{2 \pi n}{\ln n}})$ should provide a good estimate ? $\endgroup$
    – reuns
    Oct 26, 2016 at 18:15

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