The Riemann prime counting function in question is the following:
\[\Pi_0(x)=\frac{1}{2}\sum_{p^k<x}\frac{1}{k}+\frac{1}{2}\sum_{p^k\leq x}\frac{1}{k}\]
So that, for any $x$ where $x$ is exactly a prime raised to $k$, this function will jump by $\frac{1}{2k}$.
His derivation goes like this:
He finds that the logarithm of the zeta function is
\[\log{\zeta(s)}=\sum{p^{-s}}+\frac{1}{2}\sum{p^{-2s}}+\frac{1}{3}\sum{p^{-3s}}+\cdots\]
Which can be rewritten using the following identity
\[p^{-ks}=s\int_{p^k}^{\infty}x^{-s-1}ds\]
as
\[\log{\zeta(s)}=s\int_1^{\infty}\Pi_0(x)x^{-s-1}dx\]
My question now is, how come this function jumps by $\frac{1}{2k}$ at each prime power $k$, it seems like this function would jump by $k$ at each prime power? What am I not getting here?
Here is his paper:
https://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf
The last expression \[\frac{\zeta(s)}{s}=\int_0^\infty\Pi(x)x^{-s-1}dx\] Can be inverted by (a modified) Mellin Inversion: \[\Pi_0(x)=\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{\zeta(s)}{s}x^{s}dx\] And the Mellin inversion can be derived from the Fourier transform which famously has plateaus at all discontinuities of a function. Meaning that a function, $f(x)$, will be transformed (as a consequence of the Fourier transform) into \[f_0(x)=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon)+f(x-\epsilon)}{2}\]