TITLE: How to prove Chebyshev's result: $\sum_{p\leq n} \frac{\log p}{p} \sim\log n $ as $n\to\infty$?
QUESTION [22 upvotes]: I saw reference to this result of Chebyshev's:
$$\sum_{p\leq n} \frac{\log p}{p} \sim \log n \text{ as }n \to \infty,$$
and its relation to the Prime Number Theorem. I'm looking into an information-theory proof by Kontoyiannis I was wondering if anyone could give me a sense for how difficult or involved the usual proof is.
I don't really need to see the whole thing in detail, more wondering about the general difficulty/complexity of the result. Thanks!

REPLY [4 votes]: This proof differs from the others in that it uses the prime number theorem with remainder term, which is perhaps overpowered for this, but gives a somewhat shorter proof.
Using Riemann-Stieltjes integration, the sum can be written as
$$
\sum_{p \leq x} {\log p \over p}
 = \sum_{n \leq x} {\log n \over n} \Big\{{\pi(n) - \pi(n-1)}\Big\}
 = \int_2^x {\log t \over t} d\pi(t),
$$
and using integration by parts gives
$$
\sum_{p \leq x} {\log p \over p}
 = {{\log t \over t} \pi(t)}\Bigg|_2^x - \int_2^x {1 - \log t \over t^2} \pi(t) dt. \tag{1}
$$
Using the simplest form of the prime number theorem, $\pi(x) \sim x/\log x$, we see that the evaluated terms before the integral are $O(1)$ as $x \to \infty$. To handle the integral itself, we will use the prime number theorem in the form
$$
\pi(x) = \operatorname{li}(x) + O\Big({x e^{-c\sqrt{\log x}}}\Big), \tag{2}
$$
together with an expansion of $\operatorname{li}(x)$. Integrating by parts twice, we have
\begin{align*}
\operatorname{li}(x)
 &= \operatorname{li}(2) + \int_2^x {dt \over \log t}
 = \operatorname{li}(2) + \bigg\{{{t \over \log t}}\bigg|_2^x + \int_2^x {dt \over \log^2 t}\bigg\} \\
 &= {x \over \log x}
  + \bigg\{{{t \over \log^2 t}}\bigg|_2^x + 2\int_2^x {dt \over \log^3 t}\bigg\} + O(1) \\
 &= {x \over \log x} + {x \over \log^2 x}
  + O\bigg({{x \over \log^3 x}}\bigg).
\end{align*}
The error term in (2) can be subsumed by this error term, and so the integral in (1) is
\begin{align*}
\int_2^x {1 - \log t \over t^2} \pi(t) dt
 &= \int_2^x {1-\log t \over t^2}
  \bigg\{{t \over \log t} + {t \over \log^2 t}
   + O\bigg({t \over \log^3 t}\bigg) \bigg\} dt \\
 &= \int_2^x \bigg\{{1 \over t \log t}
   + {1 \over t \log^2 t}
   + O\bigg({1 \over t \log^3 t}\bigg) \bigg\} dt \\
  &\qquad- \int_2^x \bigg\{{{1 \over t}
   + {1 \over t \log t}
   + O\bigg({1 \over t \log^2 t}}\bigg)\bigg\} dt,
\end{align*}
which can be directly evaluated by the fundamental theorem of calculus as
\begin{align*}
 &= \log \log x - {1 \over \log x} - O\bigg({{1 \over 2 \log^2 x}}\bigg)
  - \log x - \log \log x + O\bigg({1 \over \log x} \bigg) \\
 &= -\log x + O(1).
\end{align*}
Plugging this back into (1) gives the desired result.