For $\displaystyle \displaystyle{\Re(s)>1} $ show $\displaystyle \displaystyle{\log\zeta(s)=s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}dx} $, where $\displaystyle \displaystyle{\pi(x)} $ is the prime counting function.
For $\displaystyle \displaystyle{\Re(s)>1} $ show $\displaystyle \displaystyle{\log\zeta(s)=s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}dx} $, where $\displaystyle \displaystyle{\pi(x)} $ is the prime counting function.
This is a problem related to analytic number theory (I believe ) , but since I know nothing about this topic , what i am going to do is only using techniques of integration ...
$\displaystyle \int_2^{\infty} \frac{ \pi(x) }{x (x^s - 1)}~dx $
$\displaystyle \int_2^3 \frac{ \pi(x) }{x (x^s - 1)}~dx + \int_3^5 \frac{ \pi(x) }{x (x^s - 1)}~dx + \int_5^7 \frac{ \pi(x) }{x (x^s - 1)}~dx + ... $
The partition is following this rule , the k-th integral has the lower limit $\displaystyle p_k$ and upper limit $\displaystyle p_{k+1} $ where $\displaystyle p_k$ denotes the k-th prime in ascending order .
Then it can be written as ( for brevity indeed ! ) :
$\displaystyle \sum_{k=1}^{\infty} \int_{p_k}^{p_{k+1}} \frac{ k}{ x(x^s - 1)}~dx $
$\displaystyle = \sum_{k=1}^{\infty} k \int_{p_k}^{p_{k+1}} \left( \frac{x^{s-1}}{x^s-1} - \frac{1}{x} \right) ~dx $
$\displaystyle = \sum_{k=1}^{\infty} k \left[ \frac{1}{s} \ln(x^s-1) - \ln(x) \right]_{p_k}^{p_{k+1}} ~dx $
$\displaystyle = \frac{1}{s} \sum_{k=1}^{\infty} k \left( \ln(1- \frac{1}{p_{k+1}^s} ) - \ln(1- \frac{1}{p_k^s} ) \right) $
$\displaystyle = \frac{1}{s} \left[ 1(a_2-a_1) + 2(a_3 - a_2) + 3(a_4 - a_3) + ... \right] $ where $\displaystyle a_k = \ln(1- \frac{1}{p_k^s} ) $
$\displaystyle = \frac{1}{s} ( -a_1 - a_2 - a_3 - a_4 - ... ) $
$\displaystyle = -\frac{1}{s} ( a_1 + a_2 + a_3 + a_4 + ... )$
$\displaystyle = - \frac{1}{s} \sum_{k=1}^{\infty} \ln(1- \frac{1}{p_k^s} ) $
$\displaystyle = - \frac{1}{s} \ln\left[ \prod_{k=1}^{\infty} \left( 1 - \frac{1}{p_k^s} \right) \right] $
$\displaystyle = - \frac{1}{s} \ln[ \frac{1}{\zeta(s)}] $
$\displaystyle = \frac{1}{s} \ln(\zeta(s)) $
Therefore , $\displaystyle \ln(\zeta(s)) = s \int_2^{\infty} \frac{ \pi(x) }{x (x^s - 1)}~dx $
The only thing you left out is the justification of switching around the terms of the sum: $\displaystyle \displaystyle{\frac{1}{s} \left[ 1(a_2-a_1) + 2(a_3 - a_2) + 3(a_4 - a_3) + ... \right]} $
but this isn't too hard to show.
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Here's a shorter way to show this but it's very similar to what you did, working backwards though.
$\displaystyle \displaystyle{\log\zeta(s)=-\sum_p \log(1-p^{-s})=-\sum_{n=2}^\infty \left[\pi(n)-\pi(n-1)\right]\log(1-n^{-s})} $.
Since $\displaystyle \displaystyle{\pi(n)\log(1-n^{s})\to0 $ as $\displaystyle n\to\infty} $, we get $\displaystyle \displaystyle{\log\zeta(s)=-\sum_{n=2}^\infty \pi(n)\left[\log(1-n^{-s})-\log(1-(n+1)^{-s})\right]} $
$\displaystyle \displaystyle{=\sum_{n=2}^\infty \pi(n)\int_n^{n+1}\frac{s}{x(x^s-1)}dx = s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}dx} $
I guess you will find another solution but much simpler .
The equation in my signature is obtained by considering the factorization of $\displaystyle x^{2n} - 1 $ which is , familiar for us : $\displaystyle = (x^2-1) [ (1- \zeta x ) (1- \zeta^2 x) ( 1- \zeta^3 x) ... ( 1- \zeta^{n-1} x ) ][ (1- \zeta^{-1} x ) (1- \zeta^{-2} x)(1- \zeta^{-3} x) ... ( 1- \zeta^{-n+1} x )] $ , $\displaystyle \zeta = exp\{ \frac{i \pi }{n} \} $
Rearrange it :
$\displaystyle = (x^2-1) [ (1- \zeta x )( 1- \zeta^{-n+1} x ) ][ (1- \zeta^2 x) ( 1 - \zeta^{-n+2} x ) ] .... [ ( 1- \zeta^{n-1} x )(1- \zeta^{-1} x )]$
then make the substitution : $\displaystyle x \mapsto e^{ix} $ and consider what will happen then :
$\displaystyle (1 - \zeta^{k}e^{ix})( 1 -\zeta^{k-n}e^{ix}) $
but $\displaystyle \zeta^{k-n} = exp\{ \frac{ik\pi}{n} - i\pi \} = - \zeta^{k}$
Therefore , $\displaystyle (1 - \zeta^{k}e^{ix})( 1 -\zeta^{k-n}e^{ix}) = (1 - \zeta^{k} e^{ix} ) ( 1 + \zeta^{k}e^{ix}) $ $\displaystyle = 1 - \zeta^{2k}e^{2ix} = - \zeta^{k}e^{ix} ( \zeta^{k}e^{ix} - \zeta^{-k}e^{-ix}) $
$\displaystyle = - exp\{ i( x + \frac{k\pi}{n} ) \} [ 2i \sin( x + \frac{k\pi}{n})] $
so we have :
$\displaystyle e^{2nix} - 1 = (e^{2ix} - 1 ) \prod_{k=1}^{n-1} \left( - exp\{ i( x + \frac{k\pi}{n} ) \} [ 2i \sin( x + \frac{k\pi}{n})] \right) $
$\displaystyle e^{inx} [ 2i\sin(nx) ] = e^{ix} [ 2i\sin(x) ] \prod_{k=1}^{n-1} \left( - exp\{ i( x + \frac{k\pi}{n} ) \} [ 2i \sin( x + \frac{k\pi}{n})] \right)$
$\displaystyle e^{inx} [ 2i\sin(nx) ] = e^{ix} [ 2i\sin(x) ] exp\{i \frac{n(n-1)\pi}{2n} \} (-1)^{n-1} 2^{n-1} e^{i(n-1)x} i^{n-1} \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n}) $
$\displaystyle \sin(nx) = 2^{n-1} \sin(x) \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n}) [ i^{n-1} (-1)^{n-1} i^{n-1} ] $
$\displaystyle \sin(nx) = 2^{n-1} \sin(x) \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n}) [ (-1)^{n-1} (-1)^{n-1} ] $
$\displaystyle \sin(nx) = 2^{n-1} \sin(x) \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n})$