I am studying analytic number theory by myself and I came across with two posts that seem to confuse me . Let $S(x)=\sum_{p\le x,\; q\le x,\; pq\gt x}\frac{1}{pq}$, where p and q are primes. Find the limit of this function.
Show that $\sum_{pq\leq x}\frac{1}{pq}$ = $(\ln \ln x)^2+O(\ln \ln x)$.
Isn't it true that $$\sum_{\begin{array}{c} p\leq x,q\leq x\\ pq>x \end{array}}\frac{1}{pq}.$$ $$ =$$ $$\sum_{p \le x}\sum_{ x/p<q\le x} \frac{1}{pq} $$ where the sums are over primes $p,q$?It has partially to do with Dirichlet's hyperbola method. So why in the first post the sum is $\pi^2/6 +Ο(1/\log x)$ while in the second it is $O(\log\log x)$? Is it that it is just a loose approximation in the second one just to solve the exercise or I am missing something? Please help.
There are two unrelated questions in the OP.
First, you ask if $$\sum_{\substack{p\leq x,q\leq x\\pq>x}}\frac{1}{pq} = \sum_{p \le x}\sum_{ x/p<q\le x} \frac{1}{pq}. \tag{1}$$ This is true.
Implicitly, you seem to be suggesting that these are related to the sums in the two linked questions are related. The sums in the two questions are
$$ \sum_{\substack{p \leq x, q \leq x \\ pq > x}} \frac{1}{pq} \tag{2} $$
and
$$ \sum_{pq \leq x} \frac{1}{pq} \tag{3}. $$
Sum $(1)$ is the same as sum $(2)$, but not the same as sum $(3)$. For example, when $x = 10$, sum $(3)$ includes the summand $\frac{1}{2 \cdot 2} = \frac{1}{4}$, but sum $(2)$ does not. More generally, sum $(3)$ includes the "large" summands coming from products of small primes, while $(2)$ omits these primes.