Asymptotics for $\sum\frac{1}{p\log p}$

I would like an asymptotic estimate for $$\displaystyle\sum_{p\leq x} \frac{1}{p\log p}$$ ($$p$$ prime).

Specifically I am trying to show that $$L^2\sum_{L^2\leq p\leq\exp(\log^2 L)}\frac{1}{p\log p}$$, where $$L:=\sqrt{\log N\log\log N}$$, is dominated by $$\log N$$ for large $$N$$.

I have tried using summation by parts but can’t get anything useful out of it. Help would be appreciated.

• If nobody answers this in a couple of hours, I’ll work it out and give you a bound. – Clayton Jan 4 '20 at 0:11

Summation by parts + the PNT $$\sum_{p\le x}1\sim \sum_{n\le x} 1/\log n$$ gives $$\sum_{p\ge x} \frac1{p\log p}\sim \sum_{n\ge x} \frac{1}{n\log^2 n} \sim \frac{1}{\log x}$$ You can replace the PNT by Mertens theorem.
In particular $$A=\sum_p \frac1{p\log p}$$ converges and $$\sum_{p\le x} \frac1{p\log p}= A-\frac{1}{\log x}(1+o(1))$$ Are you asking if there is an alternative expression for $$A$$ ?