Confused about the derivation of the identity: $\ln(\zeta(s))=s\int_{0} ^ {\infty} J(x)x^{-s-1}dx$ and it's relationship to Stieltjes integrals In the book Prime Obsession by John Derbyshire the following identity is explained:
$$\ln(\zeta(s))=s\int_{0} ^ {\infty} J(x)x^{-s-1}dx\tag{1}$$
Where J(x) is the Riemann Prime Counting Function. For reference the derivation is summarized here on page 40. In short $\ln(\zeta(s))$ is written like so:
$$\ln(\zeta(s))=s\sum_p\sum_n\frac{1}{n}\int_{p^n}^\infty{x^{-s-1}}dx\tag{2}$$
Each integral in the sum (apparently?) corresponds to a strip under $J(x)$ like so:
$J(x)$ with strips" />
I am having trouble understanding the motivation behind why one would think of the sum as related to the area under $J(x)$. It feels very nonintuitive, especially since normally the area would diverge. Derbyshire says that the $x^{-s-1}$ term in the integral allows the area of each strip to 'decrease' as x goes to infinity, allowing it to converge. I actually think the idea of area tapering off makes more sense as a double integral (A normal integral of $J(x)$ would make it so that each strip has height 1[in the z direction], but using a double integral allows the volume to taper off):
$$\ln(\zeta(s))=s\int_{0} ^ {\infty} \int_0 ^ {J(x)}x^{-s-1}dydx\tag{3}$$
Perhaps obtaining $(2)$ is a matter of simply changing $(3)$ into a $dxdy$ integration, but figuring out what exactly goes inside the $dy$ integral (and getting the desired sum over the primes and naturals) seems hard to reason out. I feel like this is caused by the unusual way $J(x)$ is defined. The idea of adding up all the strips almost makes sense, but I am having trouble connecting the dots. Tentatively can the idea of adding up the strips/rectangles like this be explained by Stieltjes integration? I am only familiar with the Riemann Integral, but if the idea of adding these strips together via. some other type of integration makes this easier to understand then I would like to know.$(1)$ is actually defined in Harold Edwards book as that type of integral:
$$\log(\zeta(s)) =\int_0^{\infty}x^{-s}dJ(x) \tag{4}  $$
Derbyshire defined the value of $J(x)$ at jump discontinuities as halfway between its new value and old value, and Edwards mentions in his book that defining the value of $J(x)$ at jump discontinuities in that manner is usual in the theory of Stieltjes integrals (p.22), which makes me believe that adding the strips and the Stieltjes integral are related in some way. That or Derbyshire added the rule about jump discontinuities for a different reason(?)
My questions are as follows:
How do you reason out $(2)$? using the double integral setup I mentioned? Is adding up the strips in this way related to Stieltjes integration? If not, could someone explain how Stieltjes integration can be used to derive $(4)$? (I am only familiar with the Riemann integral)
 A: Assuming $(n\land k)\in \mathbb{N}\land p\in \mathbb{P}$, $\log\zeta(s)$ can be derived as follows:

(1) $\quad J(x)=\sum\limits_{n=1}^x a(n)\ ,\quad a(n)= \left\{ \begin{array}{cc}
 \frac{1}{k} & n=p^k \\
 0 & \text{otherwise} \\
\end{array}\right.$

(2a) $\quad J_o(x)=\underset{\epsilon\to 0}{\text{lim}}\left(\frac{J(x-\epsilon)+J(x+\epsilon)}{2}\right)$
(2b) $\quad J_o(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\ \theta(x-n)\right)$

(3a) $\quad \log\zeta(s)=s\int\limits_0^\infty J_o(x)\ x^{-s-1}\,dx\ ,\quad\Re(s)>1$
(3b) $\quad \log\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\left(s\int\limits_0^\infty \theta(x-n)\ x^{-s-1}\,dx\right)\right),\quad\Re(s)>1$
(3c) $\quad \log\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\ n^{-s}\right),\quad\Re(s)>1$

Also $\log\zeta(s)$ can also be derived from $J_o'(x)$ as follows.

(4) $\quad J_o'(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\ \delta(x-n)\right)$

(5a) $\quad \log\zeta(s)=\int\limits_0^\infty J_o'(x)\ x^{-s}\,dx\ ,\quad\Re(s)>1$
(5b) $\quad \log\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\left(\int\limits_0^\infty \delta(x-n)\ x^{-s}\,dx\right)\right),\quad\Re(s)>1$
(5c) $\quad \log\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N a(n)\ n^{-s}\right),\quad\Re(s)>1$

Formulas (3a) and (5a) above (corresponding to formulas (1) and (4) in the OP's question above) can be shown to be equivalent via integration by parts as defined in formula (6) below. Formula (5a) can be derived from formula (3a) by setting $u=J_o(x)$ and $dv=s\ x^{-s-1}\ dx$, and formula (3a) can be derived from formula (5a) by setting $u=x^{-s}$ and $dv=J_o'(x)\ dx$.

(6) $\quad\int_0^\infty u\ dv=\left.\left(u\ v\right)\right|_0^\infty-\int_0^{\infty } v\ du$
