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The Prime Number Theorem states $\pi(n)\sim \dfrac{n}{\ln n}$.

Would there be an equally simple expression if Riemann's Hypothesis were proved true?

From Chebyshev Function, would $\pi(n)\sim \dfrac{n}{\ln n} + \sqrt n\ln n$ work?

Addendum : A relevant link : https://mathoverflow.net/questions/70713/error-term-of-the-prime-number-theorem-and-the-riemann-hypothesis

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  • $\begingroup$ jstor.org/stable/2005976?seq=1#page_scan_tab_contents $\endgroup$ – Balarka Sen Jan 25 '16 at 11:13
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    $\begingroup$ I'm not sure if I understand correctly what you mean. $\pi(n)\sim n/\ln n + \sqrt n/\ln n$ is true unconditionally, because $n/\ln n + \sqrt n/\ln n\sim n/\ln n$. $\endgroup$ – Wojowu Jan 26 '16 at 20:26
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    $\begingroup$ One idea would be to consider the upper bound for the error term $\pi(n)-n/\ln n$. Is that closer to what you mean? If so, this is relevant. $\endgroup$ – Wojowu Jan 26 '16 at 20:31
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    $\begingroup$ @Wojowu; i think i want to use pi(x) ~ x/logx + x/(logx)^2 + o() from your link, but as stated previously the asymptotes don't work $\endgroup$ – JMP Jan 26 '16 at 21:01
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    $\begingroup$ if the RH was true, the residual would be $\pi(x) - x/\ln x = \mathcal{o}(x^{1/2+\epsilon})$ for all $\epsilon > 0$. the residual would be $\pi(x) - x/\ln x = \mathcal{O}(x^{\sigma_0} / \ln x)$ if $\zeta(s)$ had a finite number of zeros at $\Re(s) = \sigma_0$ (if the RH was false) $\endgroup$ – reuns Jan 26 '16 at 22:32
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Yes, if the RH were proved true, then the error term for $\pi(x)$ in terms of $Li(x)$ would be optimal, namely $$ | \pi(x) - Li(x) | = O(\sqrt{x}\log{x}). $$ But since we can relate $\frac{x}{\log(x)}$ with $Li(x)$, we would also obtain a version with $\frac{x}{\log(x)}$. We have $$ {\rm Li} (x) - {x\over \log x} = O \left( {x\over \log^2 x} \right) \; . $$ Formulated differently, PNT only gives $$ \pi(x)={\rm Li} (x) + O \left(x \mathrm{e}^{-a\sqrt{\log x}}\right) \quad\text{as } x \to \infty $$ for some constant $a>0$, whereas with RH we get even $$ \pi(x) = {\rm Li} (x) + O\left(\sqrt x \log x\right). $$

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  • $\begingroup$ What would this "version with $x/\log x$" actually be, though? $\endgroup$ – Wojowu Jan 26 '16 at 20:31
  • $\begingroup$ The series at the end isn't convergent for any $x$. $\endgroup$ – Wojowu Jan 26 '16 at 20:33
  • $\begingroup$ What do you mean take the first terms? First how many? Isn't that an arbitrary decision to make since the terms fo the series get larger and larger? $\endgroup$ – MCT Jan 26 '16 at 20:37
  • $\begingroup$ What is the $a$ in the $\mathcal O(xe^{-a\sqrt{\log x}})$? $\endgroup$ – user153330 Jan 26 '16 at 20:47
  • $\begingroup$ @Soke: The first terms, see the answer here. $\endgroup$ – Dietrich Burde Jan 26 '16 at 20:54

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