Asymptotic of prime number pairs

Question

I am not an expert in number theory, so if someone might know the answer or a lead for the following, it would be greatly appreciated. Let $\mathcal{P}$ be the primes, and let $x$ stand for a natural number. What is the large $x$ behavior of $$\# \{ (p, q) : p, q \in \mathcal{P}, \, p < q \, , \, p \, q \leq x \},$$ where $\#$ is the number of elements of a set, as usual? In other words, how many (unordered) pairs of primes with product at most $x$ are there as $x \rightarrow \infty$? For example, it is easy to see that, asymptotically, the number should be $$\sum_{p \in \mathcal{P}, \, p \leq \sqrt{x}}\left( \frac{x/p}{\log x/p} - \frac{p}{\log p} \right),$$but, of course, the question is whether more meaning can be given to this sum, or, written as an expression involving only $x$. Thanks.

Answer 1

$$\sum_{p<q,pq\le x} 1 =\sum_{p\le \sqrt{x}} \pi(x/p)-\pi(p)$$

$$\sum_{p\le \sqrt{x}} \pi(x/p)=\sum_{p\le x/2} \frac{x/p}{\log x/p}+O(\frac{x/p}{\log^2 x/p})$$ (I used the PNT) $$=\sum_{p\le \sqrt{x}}\frac{x/p}{\log x}(1+O(\frac{\log p}{\log x}))+O(\frac{x/p}{\log^2 x}(1+\frac{\log p}{\log x}))$$

$$=\sum_{p\le \sqrt{x}}\frac{x/p}{\log x}+O(\frac{x \log p}{p\log^2 x})= \frac{x \log\log x}{\log x}+O(\frac{x}{\log x}) $$ (Mertens theorem)

$$\sum_{p\le \sqrt{x}} \pi(p) = O(\sum_{n\le \sqrt{x}} \frac{n}{\log n}) = O(\frac{x}{\log x})$$