# Math Help - Log of zeta

1. ## Log of zeta

For $\displaystyle{\Re(s)>1}$ show $\displaystyle{\log\zeta(s)=s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}dx}$, where $\displaystyle{\pi(x)}$ is the prime counting function.

2. 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 ...

$\int_2^{\infty} \frac{ \pi(x) }{x (x^s - 1)}~dx$

$\int_2^3 \frac{ \pi(x) }{x (x^s - 1)}~dx + \int_3^5 \frac{ \pi(x) }{x (x^s - 1)}~dx + \int_5^7 \frac{ \pi(x) }{x (x^s - 1)}~dx + ...$

The partition is following this rule , the k-th integral has the lower limit $p_k$ and upper limit $p_{k+1}$ where $p_k$ denotes the k-th prime in ascending order .

Then it can be written as ( for brevity indeed ! ) :

$\sum_{k=1}^{\infty} \int_{p_k}^{p_{k+1}} \frac{ k}{ x(x^s - 1)}~dx$

$= \sum_{k=1}^{\infty} k \int_{p_k}^{p_{k+1}} \left( \frac{x^{s-1}}{x^s-1} - \frac{1}{x} \right) ~dx$

$= \sum_{k=1}^{\infty} k \left[ \frac{1}{s} \ln(x^s-1) - \ln(x) \right]_{p_k}^{p_{k+1}} ~dx$

$= \frac{1}{s} \sum_{k=1}^{\infty} k \left( \ln(1- \frac{1}{p_{k+1}^s} ) - \ln(1- \frac{1}{p_k^s} ) \right)$

$= \frac{1}{s} \left[ 1(a_2-a_1) + 2(a_3 - a_2) + 3(a_4 - a_3) + ... \right]$ where $a_k = \ln(1- \frac{1}{p_k^s} )$

$= \frac{1}{s} ( -a_1 - a_2 - a_3 - a_4 - ... )$

$= -\frac{1}{s} ( a_1 + a_2 + a_3 + a_4 + ... )$

$= - \frac{1}{s} \sum_{k=1}^{\infty} \ln(1- \frac{1}{p_k^s} )$

$= - \frac{1}{s} \ln\left[ \prod_{k=1}^{\infty} \left( 1 - \frac{1}{p_k^s} \right) \right]$

$= - \frac{1}{s} \ln[ \frac{1}{\zeta(s)}]$

$= \frac{1}{s} \ln(\zeta(s))$

Therefore , $\ln(\zeta(s)) = s \int_2^{\infty} \frac{ \pi(x) }{x (x^s - 1)}~dx$

3. The only thing you left out is the justification of switching around the terms of the sum: $\displaystyle{\frac{1}{s} \left[ 1(a_2-a_1) + 2(a_3 - a_2) + 3(a_4 - a_3) + ... \right]}$

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.

$\displaystyle{\log\zeta(s)=-\sum_p \log(1-p^{-s})=-\sum_{n=2}^\infty \left[\pi(n)-\pi(n-1)\right]\log(1-n^{-s})}$.

Since $\displaystyle{\pi(n)\log(1-n^{s})\to0$ as $n\to\infty}$, we get $\displaystyle{\log\zeta(s)=-\sum_{n=2}^\infty \pi(n)\left[\log(1-n^{-s})-\log(1-(n+1)^{-s})\right]}$

$\displaystyle{=\sum_{n=2}^\infty \pi(n)\int_n^{n+1}\frac{s}{x(x^s-1)}dx = s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}dx}$

4. Hey, I was just wondering if I could see your proof of the equation in your tag.

5. Originally Posted by Chris11
Hey, I was just wondering if I could see your proof of the equation in your tag.
Equation in my tag?

6. simple pendulum. Sorry.

7. Originally Posted by Chris11
simple pendulum. Sorry.

I guess you are not going to apologize to me , right ?

Do you mean the derivation of the equation in my signature or in that post ?

8. Lol. No, I wasn't apologizing to you. Yeah, I mean the derivation of the equation in your signature.

9. Originally Posted by Chris11
Lol. No, I wasn't apologizing to you. Yeah, I mean the derivation of the equation in your signature.
I guess you will find another solution but much simpler .

The equation in my signature is obtained by considering the factorization of $x^{2n} - 1$ which is , familiar for us : $= (x^2-1) [ (1- \zeta x ) (1- \zeta^2 x) ( 1- \zeta^3 x) ... ( 1- \zeta^{n-1} x ) ][ (1- \zeta^{-1} x ) (1- \zeta^{-2} x)(1- \zeta^{-3} x) ... ( 1- \zeta^{-n+1} x )]$ , $\zeta = exp\{ \frac{i \pi }{n} \}$

Rearrange it :

$= (x^2-1) [ (1- \zeta x )( 1- \zeta^{-n+1} x ) ][ (1- \zeta^2 x) ( 1 - \zeta^{-n+2} x ) ] .... [ ( 1- \zeta^{n-1} x )(1- \zeta^{-1} x )]$

then make the substitution : $x \mapsto e^{ix}$ and consider what will happen then :

$(1 - \zeta^{k}e^{ix})( 1 -\zeta^{k-n}e^{ix})$

but $\zeta^{k-n} = exp\{ \frac{ik\pi}{n} - i\pi \} = - \zeta^{k}$

Therefore , $(1 - \zeta^{k}e^{ix})( 1 -\zeta^{k-n}e^{ix}) = (1 - \zeta^{k} e^{ix} ) ( 1 + \zeta^{k}e^{ix})$ $= 1 - \zeta^{2k}e^{2ix} = - \zeta^{k}e^{ix} ( \zeta^{k}e^{ix} - \zeta^{-k}e^{-ix})$

$= - exp\{ i( x + \frac{k\pi}{n} ) \} [ 2i \sin( x + \frac{k\pi}{n})]$

so we have :

$e^{2nix} - 1 = (e^{2ix} - 1 ) \prod_{k=1}^{n-1} \left( - exp\{ i( x + \frac{k\pi}{n} ) \} [ 2i \sin( x + \frac{k\pi}{n})] \right)$

$e^{inx} [ 2i\sin(nx) ] = e^{ix} [ 2i\sin(x) ] \prod_{k=1}^{n-1} \left( - exp\{ i( x + \frac{k\pi}{n} ) \} [ 2i \sin( x + \frac{k\pi}{n})] \right)$

$e^{inx} [ 2i\sin(nx) ] = e^{ix} [ 2i\sin(x) ] exp\{i \frac{n(n-1)\pi}{2n} \} (-1)^{n-1} 2^{n-1} e^{i(n-1)x} i^{n-1} \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n})$

$\sin(nx) = 2^{n-1} \sin(x) \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n}) [ i^{n-1} (-1)^{n-1} i^{n-1} ]$

$\sin(nx) = 2^{n-1} \sin(x) \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n}) [ (-1)^{n-1} (-1)^{n-1} ]$

$\sin(nx) = 2^{n-1} \sin(x) \prod_{k=1}^{n-1}\sin( x + \frac{k\pi}{n})$

10. That's a neat derivation. Anyways, I don't think that I'll be able to find a simpaler one, but I'll try! That's part of what math is about, right--simplicity?