Text in density of primes explanation. I was reading Riemann's Zeta Function by H. M. Edwards and I had difficulties understanding this:  
"
 \begin{equation*}(2') \quad \sum_{p<x}{1/p} \sim \log{(\log{x})}\qquad(x\to\infty),
\end{equation*}  
(...) Now 
\begin{equation*} \log{(\log{x})} = \int_{1}^{\log{x}}{\frac{du}{u}}=\int_{e}^{x}{\frac{dv}{v\cdot\log{v}}}
\end{equation*}
so (2') says that the integral of $1/v$ relative to the measure $dv/\log{v}$ diverges in the same way as the integral of $1/v$ relative to the point measure which assigns weight 1 to primes and weight 0 to all other points. In this sense (2')  can be regarded as saying that the density of primes is roughly $1/\log{v}$. "
Could you 'translate' the text to a more simple language, and how did he conclude that about the primes density?
Thank you
 A: First, he used the chain rule twice.  
Recall that when you take the derivative of a composite function, that in this case, $u = \log(x) \implies \log(u) = \log(\log(x))$.  Taking the derivative or finding the equivalent integral, $\log(u) = \int \frac{du}{u}$, and the composition on the bounds taking the integral from terms of $u$ to terms of $x$ is precisely $u = \log(x)$.
Because we are looking for a function that approximates the behavior of natural numbers, the appropriate lower bound should be 1, and if $u$ were the upper bound, then the expression would evaluate to $\log(u)$, but $u = \log(x)$ and as a result this becomes the upper bound, such that the integral arrives at the expression $\log(\log(x)) + 0$.  However this does not explain how he arrived at the expression in terms of $v$, or his application of the Prime Number Theorem
$$ \int^{\phi(a)}_{\phi(b)} f(t)dt = \int^a_b f(\phi(x))\phi'(x)dx $$
This is from Wade's Introduction to Analysis, S\3.3 -- Fundamental Theorem of Calculus, Th. 3-15 "Change of Variables"
$$\int^{log(x)}_1 \frac{du}{u}$$
Applied here, $\phi(x) = \log(x) \implies \phi'(u) = \frac{1}{u} $ and $ \phi^{-1}(u) = e^u$.
Let $u(v) = v \log v$.  Then $f = 1$, $f(\phi(u(v))) = 1$ and $\phi'(u(v)) = \frac{1}{v log(v)}$ so that
$$ \int^{x}_{e} f(\phi(u(v))) \phi'(u(v)) dv = \int^{x}_{e} \frac{dv}{v \log(v)}$$
Briefly, the Prime Number Theorem explains that the number of primes in an interval bounded by $n$ is convergent to $n / \log(n)$, so that when we divide out $n$ the fraction on average is $1/log(n)$, the density of the primes.  The sum of the inverses of all natural numbers, which I do believe he introduces first, is $log(n)$.  Euler used the product formula and bounding to determine that the sum of the inverses of the primes is convergent to $log(log(n))$ and you can read it for yourself here.
Hope that helps.
