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Feb 7, 2023 at 14:01 comment added GH from MO @KConrad Thank you. My impression is that Dimitrov's argument is essentially the same as what the proof of Axer's theorem would give in this instance. I might be wrong of course.
Feb 7, 2023 at 8:39 comment added KConrad An approach to showing PNT implies $\sum_{p \leq x} (\log p)/p = \log x + C + o(1)$, or more precisely $\sum_{n \leq x} \Lambda(n)/n = \log x - \gamma + o(1)$, without using Axer's theorem is indicated in the second half of Dimitrov's answer at mathoverflow.net/questions/95743/….
Dec 27, 2022 at 11:43 vote accept Daniel Loughran
Dec 23, 2022 at 22:49 comment added GH from MO @DanielLoughran I don't, but I have not thought about this very hard. I just recall your original question was asked from me before (in private), and I found it rather nontrivial: I ended up looking it up in Montgomery-Vaughan. BTW if you assume the PNT with a goodish error term, it is easier to derive a version of $(\ast)$ with a goodish error term. So whatever you are looking for, working with effective error terms (rather than just $o(1)$ type error terms) might be helpful in achieving your goal.
Dec 23, 2022 at 17:46 comment added Daniel Loughran Many thanks. Do you know any more direct approach which avoids using logarithmic weights?
Dec 23, 2022 at 12:03 comment added GH from MO @VladimirDotsenko Corrected, thanks. I copied the display from the original post, hence the mistake. (Now the original post is corrected as well.)
Dec 23, 2022 at 12:03 history edited GH from MO CC BY-SA 4.0
added 21 characters in body
Dec 23, 2022 at 11:59 comment added Vladimir Dotsenko I sincerely hope that you and the OP meant something like $o(1/\log x)$. In particular, $O(\log(x))$ makes the presence of $\log\log x+M$ in (*) completely pointless?
Dec 23, 2022 at 11:35 history answered GH from MO CC BY-SA 4.0