This is an application of the Shapiro -Tauberian theorem. Please see Tom Apostol's "Introduction to Analytic Number theory'' text.
Theorem: Let $\{a(n)\}$ be a non negative sequence such that
- $\displaystyle\sum\limits_{n \leq x }a(n) \biggl[\frac{x}{n}\biggr] = x\log{x} + \mathcal{O}(x)$ for all $x \geq 1$.
Then
For $x \geq 1$ we have $\sum\limits_{ n \leq x} \frac{a(n)}{n} = \log{x} + \mathcal{O}(1)$
There is a constant $A >0$ and an $x_{0} > 0$ such that $\sum\limits_{ n \leq x} a(n) \leq Bx$ for all $x \geq 1$.
There is a constant $A>0$ and an $x_{0}>0$ such that $\sum\limits_{ n \leq x} a(n) \geq Ax$ for all $x \geq x_{0}$.
Lemma 1. We have $\sum\limits_{n \leq x} \Lambda(n) \biggl[\frac{x}{n}\biggr] = \log[x]!$ where $\Lambda(n)$ denotes the Von-Mangoldt function which is defined as $\Lambda(n) = \biggl\{ \begin{array}{cc} \log{p} & \text{if} \ n=p^{m} \ \text{for some prime} \ p \\\ 0 & \text{otherwise}\end{array}$
Proof. \begin{align*} \sum\limits_{n \leq x}\Lambda(n) \biggl[\frac{x}{n}\biggr] =\sum\limits_{n \leq x}\sum\limits_{d \mid n} \Lambda(d)=\sum\limits_{n \leq x } \log{n} = \log[x]! \quad \biggl[ \because \sum\limits_{d \mid n} \Lambda(d) = \log{n} \ \biggr] \end{align*}
Lemma 2. $\displaystyle \sum\limits_{n \leq x} \Lambda(n)\biggl[\frac{x}{n}\biggr] = x\log{x} + \mathcal{O}(\log{x})$.
Proof. Taking $f(t) = \log{t}$ in the Euler's Summation formula, we have \begin{align*} \sum\limits_{n \leq x} \log{n} &= \int\limits_{1}^{x} \log{t} \ dt + \int\limits_{1}^{x} \frac{t-[t]}{t}\ dt - ( x -[x])\cdot \log{x} \\\ &= x\log{x} - x + 1 + \int\limits_{1}^{x} \frac{t-[t]}{t} \ dt + \mathcal{O}(\log{x}) \\\ &= x \log{x} + \mathcal{O}(\log{x}) \qquad \Biggl[ \because \int\limits_{1}^{t} \frac{t-[t]}{t} \ dt = \mathcal{O}\biggl(\int\limits_{1}^{t} \frac{1}{t} \ dt\biggr)=\mathcal{O}(\log{x})\Biggr] \end{align*}
Corollary. Take $a(n)= \Lambda(n)$. By Shapiro-Tauberian theorem, we then have $\sum\limits_{n \leq x} \frac{\Lambda(n)}{n} = \log{x} + \mathcal{O}(1)$
Lemma 3. For $x \geq 2$ we have $\sum\limits_{p \leq x}\biggl[\frac{x}{p}\biggr]\log{p} = x\log{x} + \mathcal{O}(x)$
Proof. Since $\Lambda(n)=0$ unless $n$ is a prime power, we have $\sum\limits_{n \leq x} \biggl[\frac{x}{n}\biggr]\Lambda(n)= \mathop{\sum\limits_{p} \sum\limits_{m=1}^{\infty}}_{p^{m} \leq x}\biggl[\frac{x}{p^{m}}\biggr] \Lambda(p^{m})$
For $p^{m} \leq x$ we have $p \leq x$. Also $\displaystyle \Bigl[\frac{x}{p^{m}}\Bigr]=0$ if $p >x$, so we can write the last sum as $\sum\limits_{p \leq x}\ \sum\limits_{m=1}^{\infty}\biggl[\frac{x}{p^{m}}\biggr]\log{p}=\sum\limits_{p \leq x} \biggl[\frac{x}{p}\biggr]\log{p} + \sum\limits_{p \leq x} \ \sum\limits_{m=2}^{\infty} \biggl[\frac{x}{p^{m}}\biggr]\log{p}$
Now we prove that the last sum above is $\mathcal{O}(x)$. We have \begin{align*} \sum\limits_{p \leq x} \log{p} \sum\limits_{m=2}^{\infty} \biggl[\frac{x}{p^{m}}\biggr] &\leq \sum\limits_{p \leq x} \log{p} \sum\limits_{m=2}^{\infty} \frac{x}{p^{m}} = x \sum\limits_{p \leq x} \log{p} \sum\limits_{m=2}^{\infty} \biggl(\frac{1}{p}\biggr)^{m} \\\ &= x \sum\limits_{p \leq x} \log{p} \cdot \frac{1}{p^{2}} \cdot \frac{1}{1-\frac{1}{p}} = x \sum\limits_{p \leq x} \frac{\log{p}}{p(p-1)}\\\ & \leq \sum\limits_{n=2}^{\infty}\frac{\log{n}}{n(n-1)} = \mathcal{O}(x) \end{align*}
Suppose you set $\Lambda_{1}(n) = \biggl\{ \begin{array}{cc} \log{p} & \text{if} \ n \ \text{is a prime} \\\ 0 & \text{otherwise}\end{array}$ then you have from the above Lemma $\sum\limits_{n \leq x} \Lambda_{1}(n) \biggl[\frac{x}{n}\biggr] = x\log{x} + \mathcal{O}(x)$ Now applying the Shapiro - Tauberian theorem with $a(n)= \Lambda_{1}(n)$ gives $\sum\limits_{p \leq x} \frac{\log{p}}{p} = \log{x} + \mathcal{O}(1)$
- Everything Can be found in Apostol's book. I just thought TeXing it here would be nice.