In H. M. Edwards' book Riemann's Zeta Function on pages $33–34$ (chapter "Riemann's paper") it reads
Of course Riemann's goal was to obtain a formula not for $J(x)$ [explained below] but for the function $\pi (x)$, that is, for the number of primes less than any given magnitude $x$. [...] Riemann inverts this relationship [...] to obtain $$\pi (x)=\sum_{n\gt 0}\frac{\mu (n)}{n}J\left(x^{\frac{1}{n}}\right),$$ [...] the series [the one above] [...] when combined with the analytical formula for $J(x)$ $$J(x)=\operatorname{li}x-\sum_{\rho}\operatorname{li} x^{\rho}-\ln 2+\int_x^\infty \frac{dt}{t\left(t^2-1\right)\ln t},\, x\gt 1,$$ it gives an analytical formula for $\pi (x)$ as desired.
(mathematical notation slightly adjusted to match current conventions, though $\pi (x)$ and $J(x)$ were not changed)
Also, on page $22$, $J(x)$ is defined as $$J(x)=\frac{1}{2}\left(\sum_{p^n\lt x}\frac{1}{n}+\sum_{p^n\le x}\frac{1}{n}\right).$$
Even though Riemann elaborated on the number of primes less than a given magnitude, in modern mathematics, $\pi (x)$ denotes the number of primes less than or equal to a given magnitude. From Riemann's Ueber die Anzahl der Primzahlen unter einer gegeben Grösse:
[...] the number of primes that are smaller than $x$ can now be determined. Let $F(x)$ be equal to this number when $x$ is not exactly equal to a prime number, but let it be greater by $\frac{1}{2}$ when $x$ is a prime number, so that, for any $x$ at which there is a jump in the value in $F(x)$, $$F(x)=\frac{F(x+0)+F(x-0)}{2}.$$ [translated to English]
(note the obsolete notation for limits; also the $F$ in $F(x)$ should have been distinguished from the $F$ in $F(x+0)$)
Now, let's change the indices to make things clear. Let $\pi (x)=\sum_{p\le x}1$ be the prime counting function and $$J(x)=\sum_{p^n\le x}\frac{1}{n},$$ $$J_0 (x)=\frac{1}{2}\left(\sum_{p^n\lt x}\frac{1}{n}+\sum_{p^n\le x}\frac{1}{n}\right)\, (\text{that's Edwards'}\, J(x)),$$ $$\pi _0 (x)=\lim_{\varepsilon\to 0}\frac{\pi (x-\varepsilon)+\pi (x+\varepsilon)}{2}\, (\text{that's Edwards'}\, \pi (x)).$$ Then we have $$\pi _0 (x)=\sum_{n\gt 0}\frac{\mu (n)}{n}J_0 \left(x^{\frac{1}{n}}\right)$$ according to this Wikipedia article (https://en.wikipedia.org/wiki/Prime-counting_function). This, however, contradicts the Mathworld article on the same topic (https://mathworld.wolfram.com/PrimeCountingFunction.html):
A modified version of the prime counting function is given by $$\begin{align}\pi _0 (p)&\equiv\begin{cases}\pi (p) &\mbox{ }\text{for}\, p\, \text{composite}\\ \pi (p)-\frac{1}{2} &\mbox{ }\text{for}\, p\, \text{prime}\end{cases}\\ \pi _0(p)&=\sum_{n=1}^\infty \frac{\mu (x)f\left(x^{1/n}\right)}{n},\end{align}$$ where $\mu (n)$ is the Möbius function and $f(x)$ is the Riemann prime counting function. [I think there should be $\mu (n)$ and $f\left(p^{1/n}\right)$ in the formula.]
The Riemann prime counting function is defined to be $$f(x)=\sum_{p^n\le x}\frac{1}{n}$$ but, looking at the formula Mathworld provides, that "$f(x)$" should be the same as $J_0 (x)$ mentioned before (which is not) and that's the contradiction. Even more confusingly, the Mathworld article on the Riemann prime counting function reads
Amazingly, the prime counting function $\pi (x)$ is related to $f(x)$ by the Möbius transform $$\pi (x)=\sum_{n=1}^\infty \frac{\mu (n)}{n}f\left(x^{1/n}\right),$$ [...] Riemann proposed that $$f(x)=\operatorname{li}(x)-\sum_{\rho}\operatorname{li}(x^{\rho})-\ln 2+\int_x^\infty \frac{dt}{t(t^2-1)},$$ [...] This formula was subsequently proved by Mangoldt.
That contradicts the Wikipedia article mentioned above which claims that $$J_0 (x)=\operatorname{li}(x)-\sum_{\rho}\operatorname{li}(x^{\rho})-\ln 2+\int_x^\infty \frac{dt}{t(t^2-1)},$$ but clearly $$\frac{1}{2}\left(\sum_{p^n\lt x}\frac{1}{n}+\sum_{p^n\le x}\frac{1}{n}\right)= \sum_{p^n\le x}\frac{1}{n}$$ is not true for every $x$.
What is true, then?
