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Math Help - Riemann's Prime Counting Function

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Riemann's Prime Counting Function

    Define  \displaystyle{J(x)=\sum_{n=1}^\infty \frac{\pi\left(x^{1/n}\right)}{n}} (note this sum is actually finite).

    1.) Show  \displaystyle{\pi(x)=\sum_{n=1}^\infty \frac{\mu(n)}{n}J\left(x^{1/n}\right)} .

    2.) Given  \displaystyle{\Re(s)>1} , show  \displaystyle{\log\zeta(s) = s\int_0^\infty J(x)x^{-s-1}dx} .
    Last edited by chiph588@; December 25th 2010 at 09:32 PM.
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  2. #2
    Super Member PaulRS's Avatar
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    1. We have: S(x) = \sum_{n\geq 1}{\tfrac{\mu(n)}{n}\cdot J\left(x^{1/n}\right)} = \sum_{n\geq 1}{\tfrac{\mu(n)}{n}\cdot \sum_{m\geq 1} \tfrac{\pi\left(x^{1/(nm)}\right)}{m}=\sum_{m, n\geq 1} \mu(n) \cdot {\left(\tfrac{\pi\left(x^{1/(nm)}\right)}{n\cdot m}\right)}

    Here note that \mu (k) goes with \tfrac{\pi\left(x^{1/n}\right)}{n} if and only if k|n. Hence we have: S(x) =\sum_{n\geq{1}} \left(\sum_{d | n}\mu(d)\right)\cdot \left(\tfrac{\pi\left(x^{1/n}\right)}{n}\right) = \tfrac{\pi\left(x^{1/1}\right)}{1} = \pi(x), since \sum_{d|n}\mu(d) = 0 for n>1, and equal to 1 for n = 1. - see the more general idea here (**)

    2.
    Here we have (*) : I(s) = \int_{2}^{\infty}{J(x) \cdot {x^{-(s+1)}}}dx= \int_2^{\infty}{\left( \sum_{n\geq 1} \tfrac{\pi\left(x^{1/n}\right)}{n\cdot x^{s+1}}\right)}dx = \sum_{n\geq 1} \int_2^{\infty} \left(\tfrac{\pi\left(x^{1/n}\right)}{n\cdot x^{s+1}}\right)}dx

    (*) the lower limit is 2 since \pi(u)= 0 for u < 2.

    Let u = x^{1/n}, then I(s) = \sum_{n\geq 1}{\int_2^{\infty}{\tfrac{\pi(u)}{u^{n\cdot s + 1}}}du} - again the lower limit will be 2 because of (*).

    Reverse the summation and the integral (since u \geq 2 this works) : I(s) = \int_2^{\infty}{\left(\tfrac{\pi(u)}{u}\cdot \sum_{n\geq 1}\tfrac{1}{u^{n\cdot s}}}\right)}du = \int_2^{\infty}{\left(\tfrac{\pi(u)}{u\cdot \left(u^s - 1\right)}\right)}du

    And the rest follows from this post.

    (**) In fact we have : \sum_{n,m\geq{1}}{f(n)\cdot g(m) \cdot h\left(n\cdot m\right)} = \sum_{n\geq{1}} \left(\sum_{d | n}f(d)\cdot g\left(\tfrac{n}{d}\right)\right)\cdot h(n) whenever we can re-arrange like that. - e.g. setting h(n) = \tfrac{1}{n^s} you see the multiplication of Dirichlet Series.
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