This is the paper I am referring to.
At the bottom of page 4, we see:
$$\log \zeta(s) = - \sum \log(1 - p^{-s}) = \sum p^{-s} + \frac{1}{2}\sum p^{-2s} + \frac{1}{3}\sum p^{-3s} + ... $$
Now, we replace
$$p^{-s} = s\int_p^\infty x^{-s-1} dx\text{ and }p^{-2s} = s\int_{p^2}^\infty x^{-s-1} dx\text{ and so on}$$
How we jump to the next step is what I do not understand.
$$\frac{\log \zeta(s)}{s} = \int_1^\infty f(x) x^{-s-1}dx$$
where:
$$f(x) = \pi(x) + \frac{1}{2}\pi(x^{\frac{1}{2}}) + \frac{1}{3}\pi(x^{\frac{1}{3}}) + ...$$
Here I use $\pi(x)$ instead of Riemann's $F(x)$.
If I were to rewrite for clarity:
$$\log \zeta(s) = -\sum \log(1 - p^{-s}) = \sum p^{-s} + \frac{1}{2}\sum p^{-2s} + \frac{1}{3}\sum p^{-3s} + ... $$
$$\frac{\log \zeta(s)}{s} = \sum\int_p^\infty x^{-s-1} dx + \frac{1}{2}\sum\int_{p^2}^\infty x^{-s-1} dx + \frac{1}{3}\sum\int_{p^3}^\infty x^{-s-1} dx + ... $$
Now, somehow, we are saying:
$$\sum\int_p^\infty x^{-s-1} dx = \int_1^\infty \pi(x) x^{-s-1}dx\text{ ---Eqn(1)}$$
$$\frac{1}{2}\sum\int_{p^2}^\infty x^{-s-1} dx = \frac{1}{2}\int_1^\infty \pi(x^\frac{1}{2}) x^{-s-1}dx\text{ ---Eqn(2)}$$
and so on
What is the basis for this? How did the sum over all primes $p$ of an integral change to just an integral and somehow the sum got absorbed into $\pi(x)$? Also, more interestingly, on one side we have sum over all primes and on RHS we have number of primes less than $x$.
Thank you
