3

I know that $$\log\zeta(s) = \sum_p \sum_n {\frac{1}{n}\,p^{-ns}}$$ And I have read that $$\log\zeta(s) = s\int_0^{\infty}J(x) \, x^{-s-1} \, dx$$ (where $J(x)$ is the function which begins at $0$ for $x = 0$ and increses by a jump of $1$ at primes $p$, by a jump of $\frac12$ at prime squares $p^2$, by a jump of $\frac13$ at prime cubes, etc; $p$ sums over all the prime numbers; the real part of $s$ is always greater than $1$; and $n$ sums over all the natural numbers greater than $0$.

But how do I get from the first to the second one?

Thanks for your help.

gen-ℤ ready to perish
  • 6,610
MaxG
  • 77

1 Answers1

1

Let us at first define the function $J$. According to your description we have $$J(x) = \sum_{p^n \leq x} \frac{1}{n},$$ and thus we get $$s \int_0^{\infty} J(x)x^{-s-1}dx = s \int_1^{\infty} J(x)x^{-s-1}dx = s\sum_{j \geq 1} \int_j^{j+1}J(x)x^{-s-1}dx$$ since $J(x) = 0$ for $x \in [0,1]$. Now try to compute the right hand side and you will get $\log(\zeta(s))$.

Con
  • 9,000
  • 1
    $\sum_{n=1}^\infty a_n n^{-s} = \sum_{n=1}^\infty a_n \int_n^\infty s x^{-s-1}dx = s\int_1^\infty (\sum_{n \le x} a_n) x^{-s-1}dx$ – reuns Jun 23 '19 at 23:25