Riemann's Prime Counting Function

1. We have: $S(x) = \sum_{n\geq 1}{\tfrac{\mu(n)}{n}\cdot J\left(x^{1/n}\right)} = \sum_{n\geq 1}{\tfrac{\mu(n)}{n}\cdot \sum_{m\geq 1} \tfrac{\pi\left(x^{1/(nm)}\right)}{m}=\sum_{m, n\geq 1} \mu(n) \cdot {\left(\tfrac{\pi\left(x^{1/(nm)}\right)}{n\cdot m}\right)}$

Here note that $\mu (k)$ goes with $\tfrac{\pi\left(x^{1/n}\right)}{n}$ if and only if $k|n$ . Hence we have: $S(x) =\sum_{n\geq{1}} \left(\sum_{d | n}\mu(d)\right)\cdot \left(\tfrac{\pi\left(x^{1/n}\right)}{n}\right) = \tfrac{\pi\left(x^{1/1}\right)}{1} = \pi(x)$ , since $\sum_{d|n}\mu(d) = 0$ for n>1, and equal to 1 for n = 1. - see the more general idea here $(**)$

2.
Here we have (*) : $I(s) = \int_{2}^{\infty}{J(x) \cdot {x^{-(s+1)}}}dx= \int_2^{\infty}{\left( \sum_{n\geq 1} \tfrac{\pi\left(x^{1/n}\right)}{n\cdot x^{s+1}}\right)}dx = \sum_{n\geq 1} \int_2^{\infty} \left(\tfrac{\pi\left(x^{1/n}\right)}{n\cdot x^{s+1}}\right)}dx$

$(*)$ the lower limit is 2 since $\pi(u)= 0$ for u < 2.

Let $u = x^{1/n}$ , then $I(s) = \sum_{n\geq 1}{\int_2^{\infty}{\tfrac{\pi(u)}{u^{n\cdot s + 1}}}du}$ - again the lower limit will be 2 because of (*).

Reverse the summation and the integral (since $u \geq 2$ this works) : $I(s) = \int_2^{\infty}{\left(\tfrac{\pi(u)}{u}\cdot \sum_{n\geq 1}\tfrac{1}{u^{n\cdot s}}}\right)}du = \int_2^{\infty}{\left(\tfrac{\pi(u)}{u\cdot \left(u^s - 1\right)}\right)}du$

And the rest follows from this post.

$(**)$ In fact we have : $\sum_{n,m\geq{1}}{f(n)\cdot g(m) \cdot h\left(n\cdot m\right)} = \sum_{n\geq{1}} \left(\sum_{d | n}f(d)\cdot g\left(\tfrac{n}{d}\right)\right)\cdot h(n)$ whenever we can re-arrange like that. - e.g. setting $h(n) = \tfrac{1}{n^s}$ you see the multiplication of Dirichlet Series.