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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 $f$


$\pi(x)=\frac{x}{\log x}+ g(x) + O(h(x))\quad$ 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.

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Yes to the second one: g(x)=Li(x)-x/log x; h(x)=sqrt(x)log x – Anthony Quas Jul 19 '11 at 7:03
Changed my post to include that $g$ must be an elementary function. – user16557 Jul 19 '11 at 7:14
and that assuming RH we can refine $\mathcal{O}(x^{1/2+\epsilon})$ to $h(x) = \mathcal{O}(\sqrt{x} \log x)$ should come from the fact that we know there are $(C T+\mathcal{O}(1)) \ln T $ non-trivial zeros $Im(\rho) < T$… – user1952009 Feb 21 at 21:21
up vote 17 down vote accepted

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 $x/\log x$.

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so x is prime when floor(pi(x))<floor(pi(x+1)) – Jon Mark Perry Jan 26 at 21:17

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