Continuation of the Riemann Prime Counting function

Question

Let $f(x)=\sum_{n=1}^{\lfloor \ln(x)\rfloor} \frac {\pi(x^{\frac 1n})}n$ ($\pi(x)$ is the Riemann prime counting function, here are some more informations). It is thanks to Riemann well known that $\pi(x) =\sum_{n=1}^\infty \frac {\mu(n)}n f(x^{\frac 1n})$ but could somebody explain to me how he prooved this? I know Möbius inversion a little bit, but I couldn't figure it out by myself how he got from the first to the second formula.

Answer 1

When everything converges, in particular if $g$ vanishes for $x < 2$ $$f(x) = \sum_{n=1}^\infty \frac1n g(x^{1/n}) \quad \iff \quad g(x) = \sum_{n=1}^\infty \frac{\mu(n)}n f(x^{1/n})$$ To convince yourself you can expand $\sum_{n=1}^\infty \frac{\mu(n)}n f(x^{1/n})= \sum_{n=1}^\infty \frac{\mu(n)}n \sum_{m=1}^\infty \frac1m g(x^{1/(nm)})$ $= \sum_{k=1}^\infty g(x^{1/k})\sum_{mn = k} \frac{\mu(n)}{n} \frac{1}{m}$ then $\sum_{mn = k} \frac{\mu(n)}{n} \frac{1}{m} = \frac1k \sum_{d | k} \mu(d)$ which is $ 0$ for $k \ge 2$