# Arbitrarily large values for $|Li(x) - \pi(x)|$

I was wondering whether there are arbitrarily large values for the $|Li(x) - \pi(x)|$. I do know that $Li(x) - \pi(x)$ changes sign infinitely often, but this does not imply that the difference stays bounded (does it?)

If an answerer could also reference me to a paper exploring this difference (especially if there is some relationship to Riemann Hypothesis), I would be grateful.

Thanks in advance.

## 3 Answers

Good lower and upper bounds of $\pi(x)-li(x)$ have been provided by Saouter and Demichel in $2010$ (not relying on RH), i.e., $$\frac{-0.2x}{\log^3 x}-\frac{12x}{\log^4 x}-C_1-C_2\le \pi(x)-li(x)\le \frac{0.51x}{\log^3 x}-C_1 \quad \forall \; x\ge 355991,$$ with constants $$C_1=li(2)-\frac{2}{\log 2}\left(1+\frac{1}{\log 2}+\frac{2}{\log^2 2}\right),$$

$$C_2=\int_2^{e^8}\frac{48}{\log^5 t}dt-\frac{24}{\log^4 2}.$$ That the difference $|\pi(x)-li(x)|$ really gets arbitrarily large had been first shown by Littlewood in 1914: assuming RH we have $$\pi(x)-li(x) = \Omega_{\pm}(x^{1/2}(\log(x)^{-1}\log \log \log x),$$ where the $\Omega_{\pm}$ means that this magnitude is reached. A detailed discussion of Littlewood's result is contained in the master thesis of Christine Lee here. Also the case that RH is false is considered.

• Is there such a result without the assumption of RH? – Linus S. Aug 13 '15 at 13:36

Assuming the Riemann Hypothesis, Helge von Koch proved in this paper that: $$\pi(x) = {\rm Li} (x) + O\left(\sqrt x \log x\right)$$

So it does not stay bounded.

• Wait, doesn't this imply that $\pi(x) - Li(x) \in O(\sqrt(x)\log x)$, implying that it does increase without bound? – Linus S. Aug 13 '15 at 13:01
• @LinusS. - misread the question, but the intent was the same :) – nbubis Aug 13 '15 at 13:05

The wiki article states that the absolute error increases without bound (second picture).