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Thread: Riemann's prime counting function

  1. #1
    Super Member
    Aug 2008

    Riemann's prime counting function

    Hi everyone. Here's something different:

    Solve the following integral via the Residue Theorem:

    f(x)=\frac{1}{2\pi i}\mathop\int\limits_{a-i\infty}^{a+i\infty} \frac{\log \zeta(s)}{s}x^s ds

    Those of you familiar with Riemann's 1859 paper know Riemann approached this integral via the xi function and infinite products. I'm quite sure it can be solved via the Residue Theorem and believe I have a solution. Anyone here interested in going over my work? It's 20 pages long and I'm not sure I have everything correct but I feel strongly the work needs to be done and made public. At present I don't have regular access to the internet so can't check this thread regularly. I'll try to check it once a day. If you guys are interested, I can post my work on the web and we could take it in steps.

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  2. #2
    MHF Contributor chiph588@'s Avatar
    Sep 2008
    Champaign, Illinois
    I'm pretty sure you can because I'm reading a book where there is a similar integral evaluated that way.

    $\displaystyle \frac{1}{2\pi i} \int_{c-\infty i }^{c+\infty i} \frac{x^s+1}{s(s+1)} \frac{d}{ds}(\log(\zeta(s)) ds = \psi_1(x) $

    where $\displaystyle \psi_1(x) = \int_{1}^{x} \psi(u) du $

    where $\displaystyle \psi(x) = \sum_{p^m \leq x} \log(p) $.
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