Problem
Prove that $$ \sum_{p \leq x} {\log p \over p} = \log x + O(1) $$ as $x \to \infty$.
Notes: $p$ ranges over primes, $\log$ is natural
Progress
Using Riemann-Stieltjes integration and integration by parts, I think you can write the sum as $$ \int_2^x {\log t \over t} d\pi(t) = O(1) - \int_2^x {1 - \log t \over t^2} \pi(t) dt. $$ From here, I tried using the prime number theorem, specifically $$ \pi(x) = \operatorname{li}(x) + O(xe^{-c\sqrt{\log x}}), $$ together with $$ \operatorname{li}(x) = {x \over \log x} + {x \over \log^2 x} + O\Big({x \over \log^3 x}\Big). $$ All the terms cancel out perfectly except for the integrals of the error terms from the prime number theorem, which diverge. If they converged, they could go into the $O(1)$ term, which would finish the proof.
Please note that this is tagged as homework.
Those integrals actually do converge. In fact, $$ O(xe^{-c\sqrt{\log x}}) = O\Big({x \over \log^3 x}\Big), $$ so there is no need to even consider them separately. I will submit this as an answer to the question this is a duplicate of now.
