I'm going through a proof of the Prime Number Theorem and the derivation of Selberg's Symmetry Formula. However, in it there is one step that is perplexing me. Would anyone be able to help explain why this is true?
$\displaystyle\sum\limits_{d^{\prime} \leq \frac{x}{d}} log^2(d^{\prime}) = log^2(\frac{x}{d}) - 2xlog(\frac{x}{d}) + 2x + O\left(\displaystyle\sum\limits_{d \leq x} log^2(\frac{x}{d})\right)$
This apparently derives from: $\displaystyle\sum\limits_{n \leq x} \frac{log(n)}{n} = \frac{1}{2}log^2(x) + C + O\left(\frac{log(x)}{x}\right)$
I don't see how. If I can get this step, then I understand the entire proof which would be awesome!
Edit: I can derive the second equation. If possible, I'm after help with the first.
Edit 2: After thinking about this for a couple of days, I've just realised an easy explanation and it doesn't require the second formula at all. You can just bound it with integrals of log^2.
Consider the sum of interest i.e., $S_x:=\sum_{n=1}^x \frac{\log(n)}n$. Note that $\frac{\log(y)}y$ for $y\geq 3$ is a monotonically decreasing function of $y$. Using this one can show that for $x\geq3$
$$S_x-S_2\geq \int_3^x dy\ \frac{\log(y)}y\geq S_{x+1}-S_3\ .$$
Carrying out the integral, we get
$$S_x-S_2\geq \frac{\log(x)^2}2- \frac{\log(3)^2}2\geq S_{x+1}-S_3 \ .$$
Thus \begin{align} S_x-S_3-\frac{\log(x)^2}2 +\frac{\log(3)^2}2&\leq \frac{\log(x-1)^2}2 -\frac{\log(x)^2}2\\ &\lesssim - \frac{\log(x)}x + O(\log x/x^2)\ , \end{align} This proves the required bound for the second relation with $C=S_3-\frac{\log(3)^2}2$.