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I was looking for representations of $\log \zeta$ and found these two:

  1. $ \displaystyle \log\zeta(s)=\color{red}{s}\sum_{n>0} \frac{P(ns)}{n\color{red}{s}}$ from here [$\color{red}{s}$ inserted by me],
  2. $ \displaystyle \log \zeta(s) = s \int_0^\infty \frac{\pi(x)}{x(x^s-1)}\,dx, $ from there.

Identify $x$ with $n$, does this somehow imply: $$ \frac{P(ns)}{\color{black}{s}}=\frac{\pi(n)}{n^s-1}\; . $$ If so, how to prove it? (Disproven here ) If not, how are 1. and 2. related?

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Have you tried plugging some values in? I don't see why it should imply that. – anon Jul 9 '12 at 20:24
No, but you're right it doesn't match. So how are (1) and (2) related? – draks ... Jul 9 '12 at 20:31
up vote 4 down vote accepted

Here's how they are related. Split $(0,\infty)$ into intervals strategically:

$$\begin{array}{c l}\int_0^\infty\frac{\pi(x)}{x(x^s-1)}dx & =\sum_{n=1}^\infty\int_{p_n}^{p_{n+1}}\frac{n}{x}\left(\sum_{k=1}^\infty x^{-ks} \right)dx \\ & =\sum_{n=1}^\infty n\sum_{k=1}^\infty\frac{p_{n+1}^{-ks}-p_n^{-ks}}{-ks} \\ & =\sum_{k=1}^\infty\sum_{n=1}^\infty \frac{1}{skp_n^{ks}} \\ & = \sum_{k=1}^\infty\frac{P(ks)}{ks} \end{array}$$

Notice the telescopy in the middle (write out some terms in the second line to see). The last line can be seen as $\log \zeta(s)$ by invoking $\zeta$'s Euler product and taking series expansions of $\log$s.

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Great answer, thanks. – draks ... Jul 9 '12 at 20:48

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