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Math Help - Derivative of the zeta function at 0

  1. #1
    Junior Member Ziaris's Avatar
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    Derivative of the zeta function at 0

    I'm not well aquainted with divergent series, so I'm just asking for help to see if this makes any sense at all. I know my final result is right, but the use of indefinite integration with divergent series leads me to question my answer. I'm pretty sure it's flawed because of that step, but here is the meat of it:

    I took the Fourier series of f(x)=e^x on arbitrary bound L, then let x\to L \to \pi\sqrt{z} to get

    \sum_{n=1}^{\infty}\frac{1}{n^2+z}=\frac{e^{\pi\sq  rt{z}}(\pi\sqrt{z}-1)+e^{-\pi\sqrt{z}}(\pi\sqrt{z}+1)}{2z(e^{\pi\sqrt{z}}-e^{-\pi\sqrt{z}})}.

    My problem is here I integrate (indefinitely) giving

    \displaystyle\sum_{n=1}^{\infty}\left[\log(n^2+z)+C\right]=\log\left(\frac{e^{2\pi\sqrt{z}}-1}{\sqrt{z}}\right)-\pi\sqrt{z}.

    Then I just figured C = 0 (my main issue) then took the limit as z went to 0 by l'Hopital's,

    \displaystyle\sum_{n=1}^{\infty}\log n = \frac{1}{2}\log 2\pi=-\zeta^{\prime}(0) by the analytic continuation of the zeta function's sum definition and its derivative, as \displaystyle\zeta^{\prime}(s)=-\sum_{n=1}^{\infty}\frac{\log n}{n^s} so of course, ignoring divergence, \displaystyle\zeta^{\prime}(0)=-\sum_{n=1}^{\infty}\log n.

    This is, quite amazingly, correct (see Wolfram eq #38), but my method of derivation seems very sketchy to me.

    My problem is is that that constant C is messing me up in the integration; is there any definite integral with certain bounds of integration that would yield the same argument, except in a (more) rigorous fashion? I can't think of any. Any ideas would be appreciated, thank you!
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  2. #2
    Junior Member Ziaris's Avatar
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    Does anyone have any experience with divergent series, and if such integration is justified?
    Last edited by mr fantastic; February 26th 2009 at 11:57 AM. Reason: Stopping the OP from shooting his/herself in the foot.
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