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Math Help - Log of zeta

  1. #1
    MHF Contributor chiph588@'s Avatar
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    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.
    Last edited by chiph588@; July 17th 2010 at 05:04 PM.
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  2. #2
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    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
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    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.

    -------

    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}
    Last edited by chiph588@; July 17th 2010 at 05:07 PM.
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  4. #4
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    Hey, I was just wondering if I could see your proof of the equation in your tag.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chris11 View Post
    Hey, I was just wondering if I could see your proof of the equation in your tag.
    Equation in my tag?
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    simple pendulum. Sorry.
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  7. #7
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    Quote Originally Posted by Chris11 View Post
    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 ?
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  8. #8
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    Lol. No, I wasn't apologizing to you. Yeah, I mean the derivation of the equation in your signature.
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  9. #9
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    Quote Originally Posted by Chris11 View Post
    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})
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  10. #10
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    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?
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