After using the Euler Product formula for the Riemann zeta function and taking the log on both sides and expanding using the Taylor series of log, this sum comes out and is fairly well known:
$$\log\zeta(s) = \displaystyle\sum_p\sum_n\frac{1}{n}p^{-ns} = \sum_p\sum_n\frac{1}{n}s\int_{p^n}^\infty x^{-s-1}dx = s\int_0^\infty J(x)x^{-s-1}dx$$
Where
$$J(x) = \displaystyle\sum_n \frac{1}{n}\pi(x^{\frac{1}{n}})$$
Where $\pi(x)$ is the prime counting function and the sums are carried out for p prime. My question is where the second integral from 0 to inf comes from and how the 2 sums got absorbed into it and somehow the inclusion of the J function.
It's a consequence of the following general fact:
To prove this, notice that Riemann-Stieltjes integration gives $$ \sum_{n\le T}{a_n\over n^s} =\int_{1^-}^T{\mathrm dA(x)\over x^s} =A(T)T^{-s}+s\int_1^T{A(x)\over x^{s+1}}\mathrm dx. $$ Since $F(s)$ converges absolutely at $s$, we know for sure that $\sum_{n\le x}|a_n|=o(x^{\Re(s)})$, which indicates that $A(T)=o(T^{\Re(s)})$. Consequently, making $T\to\infty$ gives the desired result.
To answer the question, we just need to notice that when $$ a_n= \begin{cases} 1/k & n=p^k, p\text{ prime},\\ 0 & \text{otherwise}, \end{cases} $$ we have $$ \sum_{n\le x}a_n=\sum_{k\ge1}\sum_{p\le x^{1/k}}\frac1k=\sum_{k\ge1}{\pi(x^{1/k})\over k}=J(x) $$ and $$ \sum_{n\ge1}{a_n\over n^s}=\sum_p\sum_{k\ge1}{1\over kp^{ks}}=\sum_p\log{1\over1-p^{-s}}=\log\zeta(s). $$