How to estimate $$\sum_{p\leqslant x}\sum_{q\leqslant x}\frac{1}{p+q}, \qquad\qquad(1)$$ where $p$, $q$ are prime numbers.

We have the Mertens' formula $$ \sum_{p\leqslant x} \frac{1}{p} = \log\log x+ B + O\left( \frac{1}{\log x} \right), $$ where $p$ is prime number, and $B=\gamma - \sum_{p} \left( \log \left( \frac{1}{1-1/p} \right) - \frac{1}{p} \right)$ is the Mertens constant, $\gamma$ is Euler constant.

I guess the main term of (1) is $\dfrac{x\log\log x}{\log x}$, but I don't prove it, Can you help me?


In fact, your sum (let's denote it $S$) satisfies $$ \frac x{2\log^2 x}\, (1+o(1)) \le S\le \frac{2x}{\log^2 x}\, (1+o(1)), \tag{$\ast$} $$ so that $x\log\log x/\log x$ cannot be the main term.

The upper bound is easy to prove observing that $$ \frac1{p+q} \le \frac1{2\sqrt{pq}} $$ by the AM-GM inequality. It follows that $$ S \le \frac12\,\sum_{p,q\le x} \frac1{\sqrt{pq}} = \frac12\left(\sum_{p\le x} \frac1{\sqrt p}\right)^2. $$ The sum in the RHS is known to satisfy $$ \sum_{p\le x} \frac1{\sqrt p} = \frac{2\sqrt x}{\log x}\,(1+o(1)). $$ Combining the last two estimates, we get the upper bound in ($\ast$).

For the lower bound, just notice that $$ S \ge \sum_{p,q\le x}\frac1{2x} = \frac{(\pi(x))^2}{2x} = \frac x{2\log^2 x}\, (1+o(1)) $$ by the prime number theorem.

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    $\begingroup$ Thank you very much! $\endgroup$ – arithmetic1 Apr 20 at 1:46
  • $\begingroup$ If you use the PNT then you need to provide the asymptotic, not only bounds @arithmetic1 $\endgroup$ – reuns Apr 20 at 7:52
  • $\begingroup$ @reuns: What do you mean "need"? I probably can, using partial summation; and I also can avoid using the PNT, but then the lower bound will be relaxed to $\Omega(x/\log^2(x))$. But the OP was asking for an estimate, and have not indicated that he does not want to use the PNT, correct? $\endgroup$ – W-t-P Apr 20 at 8:00
  • $\begingroup$ I find $\sum_{p\leqslant x}\sum_{q\leqslant x}\frac{1}{p+q}\sim\sum_{2\le n \le x}\sum_{2\le m \le x} \frac{1}{n+m} \frac{1}{\ln n\ \ln m}\sim \sum_{ n \le x}\sum_{ m \le x} \frac{1}{n+m} \frac{1}{\ln^2 x}$ $\sim \frac{1}{\ln^2 x}\sum_{l \le 2x} \frac{\min(l,x)}{l} \sim \frac{x (1+\ln 2)}{\ln^2 x}$ $\endgroup$ – reuns Apr 20 at 8:08
  • $\begingroup$ @reuns: Working out carefully the technical details can be quite messy, but if you are ready, you may consider adding another answer. $\endgroup$ – W-t-P Apr 20 at 8:17

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