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I asked this question before and got the desired answer in a comment, but now i want to generalize the question asked before.

given that $\pi(x) = \sum \limits_{p \leq x} 1$ is the prime counting function and $li(x) = \int \limits_{2}^{x} \frac{1}{\ln(t)}dt$ is the log integral function and $p_n$ is the $n$-th prime number and $li^{-1}(x)$ is the inverse logarithmic integral function.

assume that $|\pi(x)-li(x)| \leq \epsilon(x)$ where $\lim \limits_{x->\infty} \frac{\epsilon(x)}{li(x)} = 0$ then what is $|p_x - li^{-1}(x)| \leq ?(x)$

what is $?(x)$

** Hint ** in previous answer $|\pi(x)-li(x)| \leq x e^{-0.3 \sqrt{\ln(x)}}$ then a good bound for $|p_x - li^{-1}(x)| \leq x \ln^2(x) e^{-0.3 \sqrt{\ln(x)}}$

Note : i need the best bound without R.H.

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    $\begingroup$ Any tight bound is doomed to depend on the shape of the zero-free region for the Riemann $\zeta$ function, hence we need to know if you are willing to assume RH or not. $\endgroup$ – Jack D'Aurizio Mar 19 '17 at 17:25
  • $\begingroup$ @ Jack D'Aurizio without R.H $\endgroup$ – Ahmad Mar 19 '17 at 17:42
  • $\begingroup$ In such a case your last line essentially provides the best possible bound, through Littlewood's or Korobov-Vinogradov's estimates for the zero-free region. $\endgroup$ – Jack D'Aurizio Mar 19 '17 at 17:48
  • $\begingroup$ @Jack D'Aurizio so you say that the best is $\ln^2(x) \epsilon(x)$ ? (that is what i understood from your comment!) $\endgroup$ – Ahmad Mar 19 '17 at 17:52
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Without RH it is possibile to prove that $$p_{n}=\textrm{Li}^{-1}\left(n\right)+O\left(n\exp\left(-c\sqrt{\log\left(n\right)}\right)\right)$$ where $c>0$, see here equation $(65)$ and the refereces $[10]$ and $[12]$. In the same paper we can see, if RH holds, that $$\left|p_{n}-\textrm{Li}^{-1}\left(n\right)\right|\leq\frac{1}{\pi}\sqrt{n}\log^{5/2}\left(n\right),\,n\geq 11.$$

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