2
$\begingroup$

As it is stated, for instance, in Wikipedia, the Riemann hypothesis is equivalent to $$ |\pi(x)-{\rm li}(x)|< \frac1{8\pi}\sqrt x\log x,\qquad \mbox{for all } x\geq 2\,657, $$ but "li" denotes there the complete logarithmic integral: $$ {\rm li}(x)=\int_0^x\frac{dt}{\log t}. $$

I have checked with Mathematica that the inequality fails for $x=2\,656$. However, what happens if the offset logarithmic integral $$ {\rm Li}(x)=\int_2^x\frac{dt}{\log t} $$ is considered instead?

I have checked that, in the range $1\leq x\leq 100\,000$, the corresponding inequality holds for $x\geq 1\,447$.

Is the Riemann hypothesis equivalent to $$ |\pi(x)-{\rm Li}(x)|< \frac1{8\pi}\sqrt x\log x,\qquad \mbox{for all } x\geq 1\,447\ ? $$ All the references I have found deal with li instead of Li.

$\endgroup$
8
  • $\begingroup$ Come on $\text{li}(x)= \text{li}(2)+\text{Li}(x)$ so what is the point of your question ? $\endgroup$
    – reuns
    Dec 8, 2017 at 0:23
  • $\begingroup$ The point is that this constant could invalidate the specific determination of the constant $1/8\pi$ or the first value of $x$ from which the bound is true. $\endgroup$
    – Carlos
    Dec 8, 2017 at 1:14
  • $\begingroup$ So what, who cares to improve $\frac1{8\pi}\sqrt x\log x+\text{li}(2)$ to $\frac1{8\pi}\sqrt x\log x$ ?.. What is interesting is the method and the arguments (about the distribution of non-trivial zeros) letting us obtain those bounds. $\endgroup$
    – reuns
    Dec 8, 2017 at 1:20
  • $\begingroup$ I am not saying that the difference is important. I am just asking whether the theorem is stated with li for any reason or the version with Li is also valid. In the second case, the number of exceptions is a bit lower. Nothing important, of course, but I think the statement becomes a bit nicer. $\endgroup$
    – Carlos
    Dec 8, 2017 at 1:26
  • $\begingroup$ In that case replace $\frac{1}{8\pi}$ by $1$, it will be even nicer, you can even choose a constant such that it is iff for $x > 1$ $\endgroup$
    – reuns
    Dec 8, 2017 at 1:32

1 Answer 1

3
$\begingroup$

$\DeclareMathOperator{\Li}{Li}$

Is the Riemann hypothesis equivalent to $$\lvert\pi(x)- \Li(x)|< \frac1{8\pi}\sqrt x\log x,\qquad \text{for all } x\geq 1\,447\ ?$$

Not quite; unless I made a silly implementation error the inequality $$\Li(x) > \pi(x) + \frac{1}{8\pi}\sqrt{x} \,\log x$$ holds for $y < x < 1451$, where $y \approx 1450.86$. The lower bound of $1\,447$ is correct if you consider only integer values of $x$. (However, if my computations are sufficiently accurate and the Riemann hypothesis is wrong, then it would be logically equivalent to the assertion that the inequality holds for all real $x \geqslant 1447$.)

But that is rather unimportant, the Riemann hypothesis is equivalent to $$\lvert \pi(x) - \Li(x)\rvert < \frac{1}{8\pi}\sqrt{x}\,\log x \qquad \text{for all } x \geqslant x_0 \tag{$\ast$}$$ with a suitable $x_0$. According to my computations, $x_0 = 1451$ works and is the smallest to work.

Even that is rather unimportant, however. There is no reason to believe that the constant $\frac{1}{8\pi}$ is significant. From skimming the papers Rosser and Schoenfeld 1975 and Schoenfeld 1976 it seems to me that Schoenfeld could have proved the analogous result with a (slightly) smaller constant (and correspondingly a larger $x_0$), but the constant $\frac{1}{8\pi}$ is what comes naturally out of the computations from R-S 1975. And $\sqrt{x}\,\log x$ may well be of too large order for a sharp bound, it could, as far as I know, be that $\pi(x) - \Li(x)$ belongs to $O(\sqrt{x})$ or even $o(\sqrt{x})$. So far, however, $O(\sqrt{x}\,\log x)$ is the best we can prove (assuming RH).

The importance of Schoenfeld's result lies in the fact that he obtained an explicit constant - and a rather smallish one - for von Koch's 1901 theorem that RH implies $\pi(x) - \Li(x) \in O(\sqrt{x}\,\log x)$.

If the constant $\frac{1}{8\pi}$ worked for $\operatorname{li}(x)$ but not for $\Li(x)$, that would be interesting because it would imply a very precise result about the growth of $\lvert\pi(x) - \Li(x)\rvert$ (under RH). If we could prove that the constant works for $\operatorname{li}(x)$ (as Schoenfeld did), but could not prove that it works for $\Li(x)$, that would be interesting too, but less so.

I am just asking whether the theorem is stated with $\operatorname{li}$ for any reason or the version with $\Li$ is also valid.

The main reason people work with $\operatorname{li}$ rather than the offset logarithmic integral is that $\operatorname{li}$ has been extensively tabled, while $\Li$ hasn't. Of course one would only have to subtract a constant ($\operatorname{li}(2)$), so it's not a huge deal, but it's more convenient to skip that if it's not really needed. Another reason may be that for $\Li$ it's the second sign change of $\pi(x) - \Li(x)$ that is of considerable interest, while it's the first sign change for $\pi(x) - \operatorname{li}(x)$.

Let's now come to the proof that Schoenfeld's result holds too when we replace $\operatorname{li}$ with $\Li$ (and then we can even pick a smaller $x_0$). Schoenfeld obtained $\lvert\pi(x) - \operatorname{li}(x)\rvert < \frac{1}{8\pi}\sqrt{x}\,\log x$ for $x \geqslant 2657$ as a corollary of Theorem 10 in Schoenfeld 1976 (the numbering of sections and theorems is continued from that of R-S 1975, thus Theorem 10 is the first theorem in that paper). The pertinent parts of that theorem are \begin{align} \lvert \vartheta(x) - x\rvert &< \frac{1}{8\pi}\sqrt{x}(\log x)^2 &&\text{if } 599 \leqslant x,\tag{1} \newline \lvert \vartheta(x) - x\rvert &< \frac{1}{8\pi}\sqrt{x}(\log x - 2)\log x &&\text{if } 23\cdot 10^8 \leqslant x. \tag{2} \end{align} I'll take these on trust here. From then on it's quite simple. Let $F$ be any primitive of $x \mapsto \frac{1}{\log x}$ on $(1, +\infty)$ and suppose that $1 < \xi \leqslant 23\cdot 10^8 \leqslant x$. First, via summation by parts we obtain \begin{align} \pi(x) - \pi(\xi) &= \frac{\vartheta(x)}{\log x} - \frac{\vartheta(\xi)}{\log \xi} + \int_{\xi}^x \frac{\vartheta(y)}{y(\log y)^2}\,dy \newline &= \frac{x}{\log x} - \frac{\xi}{\log \xi} + \int_{\xi}^x \frac{dy}{(\log y)^2} \newline &\quad+ \frac{\vartheta(x) - x}{\log x} - \frac{\vartheta(\xi) - \xi}{\log \xi} + \int_{\xi}^x \frac{\vartheta(y) - y}{y(\log y)^2}\,dy \newline &= F(x) - F(\xi) + \frac{\vartheta(x) - x}{\log x} - \frac{\vartheta(\xi) - \xi}{\log \xi} + \int_{\xi}^x \frac{\vartheta(y) - y}{y(\log y)^2}\,dy. \end{align} Now one can rearrange and insert a couple of absolute values to obtain an inequality. If we now suppose $\xi \geqslant 599$, by $(1)$ and $(2)$ we obtain

\begin{align} \lvert \pi(x) - F(x)\rvert &\leqslant \frac{\lvert \vartheta(x) - x\rvert}{\log x} + \Bigl\lvert \underbrace{\pi(\xi) - F(\xi) - \frac{\vartheta(\xi) - \xi}{\log \xi} }_{\xi'} \Bigr\rvert + \int_{\xi}^x \frac{\lvert \vartheta(y) - y\rvert}{y(\log y)^2}\,dy \newline &< \frac{\sqrt{x}(\log x - 2)}{8\pi} + \lvert\xi'\rvert + \frac{2(\sqrt{x} - \sqrt{\xi})}{8\pi} \\ &= \frac{1}{8\pi}\sqrt{x}\,\log x + \lvert\xi'\rvert - \frac{\sqrt{\xi}}{4\pi}. \end{align}

Now it remains to see if for our chosen $F$ - Schoenfeld used $\operatorname{li}$, you are interested in $\Li$ - we can find a $\xi \in [599, 23\cdot 10^8]$ such that $$\lvert \xi'\rvert \leqslant \frac{\sqrt{\xi}}{4\pi},$$ which yields $\lvert \pi(x) - F(x)\rvert < \frac{1}{8\pi}\sqrt{x}\,\log x$ for $x \geqslant 23\cdot 10^8$, and then to check how low we can take $x_0$ so that the inequality holds for all $x \geqslant x_0$. Schoenfeld used $\xi = 10^8$ (lots of smaller [or larger] $\xi$ also work) and found $\lvert \xi'\rvert \approx 88.26$ with $F = \operatorname{li}$. For $F = \Li$ and $\xi = 10^8$ we obtain $\lvert\xi'\rvert \approx 87.23$. These values are much smaller than $\frac{\sqrt{\xi}}{4\pi} = \frac{2500}{\pi} \approx 795.775$, thus we see there's a loot of room to choose our favourite $F$ from.

We now have established $(\ast)$ for $x \geqslant 23\cdot 10^8$. Schoenfeld referred to unpublished tables and older results (e.g. from Rosser and Schoenfeld 1962) to establish the inequality for smaller $x$. Nowadays it's probably simplest to do a brute-force check. With the current computing power (of even a mobile phone) no advanced programming knowledge [but a little knowledge of not too inefficient algorithms] or optimisation is required to run the check for $x < 23\cdot 10^8$ in a matter of seconds.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .