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As the question in the title asks, does $\pi(x) \geq \mathrm{Li}(x)$ imply that $\vartheta(x) \geq x$? Here $\pi(x) = \#\{p \leq x\}$, $\vartheta(x) = \sum_{p \leq x} \log p$ and $\mathrm{Li}(x) = \int_{2}^{x} \frac{dt}{\log t}$.

Alternatively, is there a number $x$ such that $\pi(x) \geq \mathrm{Li}(x)$ and $\vartheta(x) \leq x$, or a number $x$ such that $\pi(x) \leq \mathrm{Li}(x)$ and $\vartheta(x) \geq x$?

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    $\begingroup$ It might help for non-experts in prime number theory to recall what $\theta(x)$ is. $\endgroup$
    – Daniel Loughran
    Commented Feb 16, 2017 at 18:12
  • $\begingroup$ My gut feeling is that the implication in the title is not true, but what do I know :P $\endgroup$
    – Wojowu
    Commented Feb 16, 2017 at 23:04
  • $\begingroup$ You can find different information about error terms from these two problems in the article S. B. Stechkin, A. Yu. Popov, “The asymptotic distribution of prime numbers on the average”, Uspekhi Mat. Nauk, 51:6(312) (1996), 21–88. See also mathoverflow.net/questions/48461/… $\endgroup$
    – Alexey Ustinov
    Commented Feb 17, 2017 at 15:01

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