On a double sum involving prime numbers 
$$\sum_{i,j=1}^{\infty}\left[\frac{x}{p_ip_j}\right]=x\sum_{p_ip_j\leq x}\frac{1}{p_ip_j}+O(x),p_i< p_j$$
  where $p_i$ is the $i$th prime, and "[ ]" represents the largest integer not exceeding...

I don't know how to deal with it. Could you give me a proof?
 A: The sum on the left is
$$
\sum_{i,j=1}^{\infty} \left[ \frac x{p_i p_j} \right] = \sum_{p_i p_j \le x} \left[ \frac{x}{p_i p_j} \right]
$$
because if $p_i p_j > x$ then the integer part is just $0$, hence you don't need to sum over infinitely many terms. It remains to see that there is $O(x)$ terms in that sum, since
$$
z = [z] + \{z\} \qquad \text{ with } \qquad \{z\} \le 1. 
$$
so that you can look at the sum on the left as the sum on your right + the remainders, which you will only need to "count" how many of them appear, since all the terms you will sum will be less than $1$. Hence showing that there is at most $O(x)$ terms in that sum shows that you can bound the remainder's sum by $O(x) \times 1 = O(x)$.
The number of terms in that sum is
$$\begin{align*}
 \# \left\{ (p_i,p_j) \, | \, p_i p_j \le x, p_i < p_j \right\} &\le \, \# \left \{ n \, | \, n \text{ is composite }, n \le x \right \} \\
&= x - \pi(x) \\
&\sim x - \frac {x}{\log x} \\
&= O(x).
\end{align*}$$
(A rough bound though would just be $x$ and it would still be $O(x)$, I just thought I'd do a little better.)
(ADDED : Aleks was wondering why my inequality held, so I'll add the proof here. Call the set with the prime couples $A$ and the set with the composites $B$. The application which takes a couple $(p_i, p_j)$ with $p_i < p_j$ to their product $p_i p_j = n$ is injective, for if $p_i^1 p_j^1 = n_1 = n_2 = p_i^2 p_j^2$, since $p_i^k < p_j^k$ we must have $p_i^1 = p_i^2$ and $p_j^1 = p_j^2$. Since this application is injective, $\# A \le \# B$. )
This gives you
$$\begin{align*}
\sum_{i,j=1}^{\infty} \left[ \frac{x}{p_i p_j} \right] &= \sum_{p_i p_j \le x} \left[ \frac x{p_i p_j} \right]\\
&= \sum_{p_i p_j \le x} \left( \frac x{p_i p_j} - \left\{ \frac{x}{p_ip_j} \right\} \right)\\
&= x \left( \sum_{p_i p_j \le x} \frac 1{p_i p_j} \right) - \sum_{p_i p_j \le x} \left\{ \frac{x}{p_ip_j} \right\}
\end{align*}$$
and since the last sum contains $O(x)$ terms less than $1$, it is $O(x)$. 
Hope that helps,
A: Since $\pi_2(x):=\#\{n \leq x:\omega(n)=2 \}\ll \frac{x}{(\log x)^2}(\log \log x)^2$, the error term is $\frac{x}{(\log x)^2}(\log \log x)^2=o(x).$ $\omega(n)$ denotes the number of primes dividing $n$.
