Introduction
There is some disagreement on what notation to use for the weighted prime function, but I'm gonna use the notation $J(x)$ for it ($f(x)$ is also used a lot). The function is defined as follows (according WolframMathWorld): $$J(x) = \sum_{n=1}^{\log_2(x)}\frac{\pi(x^{\frac{1}{n}})}{n}$$ Here $\pi(x)$ is the prime counting function. Note that we put the end to the series at $\log_2(x)$ because after that the terms become $0$. Intuitively, the function is essentially "counting prime powers $p^n$ with a weight of $\frac{1}{n}$". So when $x$ is a prime the function "jumps" by $1$, when $x$ is a square of a prime it jumps by $\frac{1}{2}$, when $x$ is a cube of a prime it jumps by $\frac{1}{3}$ and so on.
This seemingly useless function satisfies the following amazing relationship:
$$\displaystyle \ln \zeta (n)=n\int _{0}^{\infty }J(x)x^{-n-1}\,\mathrm {d} x$$
Now I'm interested in all kinds of series involving $\zeta(n)$ so this got me thinking; I have just been discussing a similar infinite series and now I thought of the following one:
$$\sum_{n=2}^{\infty} \frac{\ln(\zeta(n))}{n}$$
You probably recognize the inside of the sum; it's just the relationship above, but we've divided by $n$. So we can rewrite this as: $$\sum_{n=2}^{\infty} \frac{\ln(\zeta(n))}{n} =\sum_{n=2}^{\infty} \int _{0}^{\infty }J(x)x^{-n-1}\,\mathrm {d} x$$
Now I'm going to simplify this:
\begin{align} \sum_{n=2}^{\infty} \frac{\ln(\zeta(n))}{n} &= \sum_{n=2}^{\infty} \int _{0}^{\infty }J(x)x^{-n-1}\,\mathrm {d} x \\\\ & = \int _{0}^{\infty}(\sum_{n=2}^{\infty} J(x)x^{-n-1})\,\mathrm {d} x \\\\ & = \int _{0}^{\infty}J(x)(\sum_{n=2}^{\infty}\frac{1}{x^{n+1}})\,\mathrm {d} x \end{align}
According to WolframAlpha the series can be simplified as follows: $$\sum_{n=2}^{\infty}\frac{1}{x^{n+1}} = \frac{1}{(x-1)x^2}$$
And thus, we arrive at the final result that:
$$\sum_{n=2}^{\infty} \frac{\ln(\zeta(n))}{n} = \int _{0}^{\infty}\frac{J(x)}{(x-1)x^2}\,\mathrm {d} x$$
Question
Is this correct? I feel like I did some mistake when I switched the sum and integral. Also note that the following $$\sum_{n=2}^{\infty}\frac{1}{x^{n+1}} = \frac{1}{(x-1)x^2}$$ only converges for $|x| > 1$ which probably makes my derivation false. Any ideas on whether it is true what I have done or not?