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Since the Riemann hypothesis is equivalent to $\pi(x) = \text{Li}(x) + O(\sqrt x \log x)$,

One would expect that a plot of $|\pi(x) - \text{Li}(x)|$ and $\sqrt x \log x$ would show $|\pi(x) - \text{Li}(x)|$ coming near $\sqrt x \log x$. For values of $x$ up to even $10^8$ this does not happen. Does anyone know when the predicted asymptotic behavior shows up in a plot?

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    $\begingroup$ No, one would expect it would come near $C \sqrt{x} \log x$ for an appropriate constant. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 19, 2011 at 19:58
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    $\begingroup$ On the Riemann Hypothesis, Schoenfeld showed the constant $C$ can be taken as $1/(8\pi)$, and (iirc) this can not be improved on. $\endgroup$
    – Stopple
    Commented Jul 19, 2011 at 20:33
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    $\begingroup$ I'm confused by all your comments above. The big-O notation means that the error term is bounded by $\sqrt{x}\log(x)$, not that it's asymptotically proportional to it. See en.wikipedia.org/wiki/Big_O_notation. $\endgroup$
    – André Henriques
    Commented Jul 19, 2011 at 20:52
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    $\begingroup$ @Paul, I just looked up the Schoenfeld result: On RH, $|\pi(x)-li(x)|$ is less than $x^{1/2}\log(x)/(8\pi)$ for $2,657\le x$, no mention that this is optimal. $\endgroup$
    – Stopple
    Commented Jul 19, 2011 at 21:31
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    $\begingroup$ Once I verified that Schoenfeld's result implies $|\pi(x)-li(x)|<x^{1/2}\log x$ for all $x>2$. $\endgroup$
    – GH from MO
    Commented Jul 19, 2011 at 22:16

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The $\log x$ factor is a result of Koch (1901) and still the best known consequence of the Riemann Hypothesis, but it is probably very far from the truth. Littlewood (1914) proved that this factor is $\Omega(\log\log\log(x)/\log x)$, while Montgomery conjectures that the truth is around $(\log\log\log x)^2/\log x$. The last information was communicated to me by Pintz around a year ago.

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    $\begingroup$ It reminds me of the quote "$\log \log x$ approaches infinity but has never been observed to do so." $\endgroup$
    – Victor Miller
    Commented Jul 19, 2011 at 21:02
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    $\begingroup$ I thought the quote was "$\log\log\log x$ goes to infinity with great dignity." $\endgroup$
    – François G. Dorais
    Commented Jul 19, 2011 at 21:41
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    $\begingroup$ Montgomery's conjecture is discussed in his paper "The zeta function and prime numbers," Proceedings of the Queen's Number Theory Conference, 1979, Queen's Univ., Kingston, Ont., 1980, 1-31. He calculates, assuming RH and the zeros of the zeta function are linearly independent, the limiting distribution of the quantity $e^{-y/2}(\psi(e^y)-e^y)$. Once one knows this distribution it is natural to conjecture how large $|e^{-y/2}(\psi(e^y)-e^y)|$ can be (infinity often) on probabilistic grounds. $\endgroup$
    – Mark Lewko
    Commented Jul 20, 2011 at 0:09
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    $\begingroup$ @GH, I temporarily put a copy up at: lewko.files.wordpress.com/2011/07/… the relevant section starts on page 14. $\endgroup$
    – Mark Lewko
    Commented Jul 20, 2011 at 2:12
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    $\begingroup$ @Mark: Thanks for the paper! @GH: Gonek has a conjecture for the summatory function of the Moebius function which is analogous to Montgomery's conjecture for the error term in PNT. The details are in Nathan Ng's paper "The distribution of the summatory function of the Moebius function" which he has posted on his web-page. $\endgroup$
    – Micah Milinovich
    Commented Jul 20, 2011 at 14:43
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The Riemann Hypothesis is also equivalent to $|\pi(x) - Li(x)| = O(x^{1/2 + \epsilon})$, so let's look at that instead. In other words, $\log$ of the error should be about $(1/2) \log x$.

The sequence of points plotted below is $( \log x,\ \log |\pi(x) - Li(x)|)$ for $x=10^k$, with $1 \leq k \leq 23$. The straight line has slope $1/2$, with constant term chosen by a least squares fit (specifically, the line is $x/2 -1.24878$). Interpreted in this way, you can definitely see the promised asymptotic behavior.

    alt text(source)

(Data set courtesy of Wikipedia)

Note: My $\log$'s are base $10$, since my data set was binned by powers of $10$ already. Of course, that doesn't effect the slope.

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