I have read that the Riemann Hypothesis is equivalent to
$\pi(x)=\text{Li}(x)+O(\sqrt{x}\log x)$
Is there an analogous statement saying the Riemann Hypothesis is equivalent to
$\pi(x)=\frac{x}{\log x}+ O(f(x))\quad$ for some $f$
or
$\pi(x)=\frac{x}{\log x}+ g(x) + O(h(x))\quad$ for some elementary function $g$ and $h$
I'm guessing that $f$ could not possibly be $\sqrt{x}\log x$ because I plotted
$\frac{\text{Li}(x)-x/\log(x)}{\sqrt x\log x}$ and it looked like it grew without bound as $x$ goes to infinity.
Best Answer
It is not hard to show that $$\mathrm{Li}(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right)$$ for any $m \geq 0$ (just use the definition of $\mathrm{Li}(x)$ and repeated integration by parts). Thus $$\pi(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right).$$ It is not possible to improve on this (this is true unconditionally; you don't even need the Riemann hypothesis). So $\mathrm{Li}(x)$ really is the "better" approximation to $\pi(x)$ compared to $x/\log x$.