# show that $\sum_{pq \leq x} \frac{\log p \log q}{pq (\log p + \log q)} = \log x + O(\log \log x)$

I've been reading through Atle Selberg's "Elementary" proof of the Prime Number Theorem. That seems to be an ironic usage of the word. A lot of discussions focus on the Selberg formula:

$$\theta(x) \;\log x + \sum_{p \leq x} \log p \; \theta\big(\frac{x}{p}\big) = 2x \log x + O(x)$$

The left side is almost suggesting to let $p=1$ be a prime number. Here $\theta$ is the sum of logs of primes:

$$\theta(x) = \sum_{p \leq x} \log p = \log \bigg[ \prod_{p \leq x} p \bigg]$$ Here emphasizing the product of all prime numbers (rather than a factorial), which can be used to prove the infinitude of primes. We can write as correlations of the sequence $\{ \log p: p \text{ prime}\}$.

$$\sum_{pq \leq x} \log p \, \log q + \sum_{p \leq x} \log^2 p = 2x \log x + O(x)$$

Let me go back a few steps and go through three steps in Selberg's derivation.

$$\sum_{p \leq x} \log p = x\log x + O(x) \tag{A}$$

This step easily implies

$$\sum_{pq \leq x} \frac{\log p \log q}{pq} = \frac{1}{2} \log^2 x + O(\log x) \tag{B}$$

and then we can do a partial summation to get :

$$\sum_{pq \leq x} \frac{\log p \log q}{pq (\log p + \log q)} = \log x + O(\log \log x) \tag{C}$$

In fact I do not see at all how (A) → (B) → (C)

I simply do not have the facility to do all the re-arrangments he talks about. Here is summation by parts:

$$\sum_{k=m}^n f_k \Delta g_k = [f_{n+1}g_{n+1}-f_mg_m]-\sum_{k=m}^n g_{k+1}f_k$$