# Riemann hypothesis and the logarithmic integral

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.

• Come on $\text{li}(x)= \text{li}(2)+\text{Li}(x)$ so what is the point of your question ? Dec 8, 2017 at 0:23
• 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. Dec 8, 2017 at 1:14
• 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. Dec 8, 2017 at 1:20
• 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. Dec 8, 2017 at 1:26
• 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 $x > 1$ Dec 8, 2017 at 1:32

$$\DeclareMathOperator{\Li}{Li}$$

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.