# How to prove this relationship between $\log\zeta(s)$ and prime counting function proposed in Riemann's paper?

This formula is considerably important in Riemann's paper "On the Number of Primes Less Than a Given Magnitude", giving the relationship between prime counting function and zeta function, but I do not get how the last line was derived: \begin{align} \log\zeta(s)&=-\sum_{p\ \rm prime} \log(1-p^s)\\ &=\sum_{p\ \rm prime}\sum_{n=1}^\infty\frac{1}{n}p^{-ns}\\ &=\sum_{p\ \rm prime}\sum_{n=1}^\infty\frac{1}{n}s\int_{p^n}^\infty x^{-s-1} dx\ \left(p^{-ns}=s\int_{p^n}^\infty x^{-s-1} dx\right)\\ &=s\int_1^\infty f(x)x^{-s-1}dx,\text{ where }f(x)=\sum_{n=1}^\infty \frac{1}{n}\pi(x^{1/n}) \end{align}

• why do you have an equality in parenthesis? Commented Jun 25, 2023 at 21:16
• it is a substitution made to the previous line according to the identity Commented Jun 25, 2023 at 21:17
• It's a trick allowing you to "extend your integral beginning from $1$ instead of $p^n$", I must go eating dinner, if it is still here after I'll answer you :) Commented Jun 25, 2023 at 21:22

The thing is: you want your integral to go from $$1$$ instead of $$p^n$$, note that we can exchange the summing since these are positive, for fixed $$n$$:
$$\sum_p \int_{p^n}^\infty s x^{-s-1} dx =\sum_p \int_{1}^\infty s x^{-s-1} 1_{x\geqslant p^n}(x) dx = \int_{1}^\infty s x^{-s-1} \sum_p 1_{x\geqslant p^n}(x) dx\\ = \int_{1}^\infty s x^{-s-1} \sum_p 1_{x^{1/n}\geqslant p}(x) dx =\int_{1}^\infty s x^{-s-1}\pi(x^{1/n}) dx$$
• I do not quite get the notation $1_{x\geq p^n} (x)$ you are using here Commented Jun 25, 2023 at 21:37
• It’s the characteristic function of the set $\{x, x \geq p^n\}$ Commented Jun 25, 2023 at 21:38
• So 1 if $x \geq p$, 0 otherwise Commented Jun 25, 2023 at 21:39