What is the Asymptotic behaviour of the following quantity ? \begin{eqnarray*} \sum_{p- \text{prime} \\ p \leq x } \frac{ \ln (p) }{p}. \end{eqnarray*}
Motivation: I am looking at this paper https://scholar.princeton.edu/sites/default/files/ashvin/files/math229xfinalproject.pdf
This has got me to think about the number of numbers that can be expressed as the product of two primes \begin{eqnarray*} \pi_{ \alpha \beta} (x) &=& \sum_{p \leq q- \text{prime} \\ pq \leq x } 1 \\ &=& \sum_{p- \text{prime} \\ p \leq x } \pi \left(\frac{x}{p} \right) \\ &=& \sum_{p- \text{prime} \\ p \leq x } \frac{x/p}{\ln( x/p)} \\ &=& \sum_{p- \text{prime} \\ p \leq x } \frac{x}{p(\ln( x)-\ln(p))}. \end{eqnarray*} Geometrically expanding \begin{eqnarray*} \frac{1}{\ln( x)-\ln(p)}= \frac{1}{\ln(x)} + \frac{\ln(p)}{(\ln(x))^2}+ \cdots. \end{eqnarray*} We have \begin{eqnarray*} \pi_{ \alpha \beta} (x) &=& \frac{x}{\ln(x)} \sum_{p- \text{prime}} \frac{1}{p} +\frac{x}{(\ln(x))^2} \sum_{p- \text{prime}} \frac{\ln(p)}{p} +\cdots \end{eqnarray*} Now the recognise first sum as the content of Merten's second theorem https://en.wikipedia.org/wiki/Mertens%27_theorems \begin{eqnarray*} \sum_{p- \text{prime}} \frac{1}{p} = m + \ln( \ln (x)) +\cdots \end{eqnarray*} where $m$ is the Merten-Meissel constant. But what about the second sum ? \begin{eqnarray*} \sum_{p- \text{prime}} \frac{\ln(p)}{p} = ? \end{eqnarray*} Ideally a good answer will tell me how to approach calculating a quantity like this & a wonderful answer will perform the calculation & indicate how to calculate higher order terms.
As usual, thanks in advance, for helpful comments & answers.