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Math Help - The sum of function series

  1. #1
    Senior Member Shanks's Avatar
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    The sum of function series

    The funcation series arise in the invastigation of a interesting problem. How to find the sum? help in need. Here it is:
    \sum \frac{x^2}{k^2+x^2}
    the sum is taken over all positive integers k.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    You can use the following result:

    If f is analytic on \mathbb{C} except by a finite number of poles z_1,\ldos,z_m ( z_j\not\in\mathbb{Z} ) and f(z)=O(z^{-2}) as z\to \infty then,

    \displaystyle \sum_{-\infty<k<+\infty}f(k)=-\pi\displaystyle\sum_{j=1}^m\textrm{res}_{z=z_j}f(  z)\cot (\pi z)

    You'll obtain S=\pi x \coth (\pi x) where x>0.
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  3. #3
    Senior Member Shanks's Avatar
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    thanks for your help. ha, complex analysis is really powerful.
    I think i got it now, I will do it again by myself. Thanks again!
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Shanks View Post
    The funcation series arise in the invastigation of a interesting problem. How to find the sum? help in need. Here it is:
    \sum \frac{x^2}{k^2+x^2}
    the sum is taken over all positive integers k.
    Alternatively You can start from the [well known...] 'infinite product'...

    \displaystyle \varphi(x)= \frac{\sinh \pi x}{\pi x} = \prod_{k=1}^{\infty} (1+\frac{x^{2}}{k^{2}}) (1)

    ... and from (1) You derive first...

    \displaystyle \ln \varphi(x) = \sum_{k=1}^{\infty} \ln (1+\frac{x^{2}}{k^{2}}) (2)

    ... and then...

    \displaystyle \frac{d}{dx} \ln \varphi(x) = 2\ \sum_{k=1}^{\infty} \frac{\frac{x}{k^{2}}}{1+\frac{x^{2}}{k^{2}}}= 2\ \sum_{k=1}^{\infty} \frac{x}{k^{2}+x^{2}} (3)

    ... so that is...

    \displaystyle \sum_{k=1}^{\infty} \frac{x^{2}}{k^{2} + x^{2}} = \frac{x}{2}\ \frac{d}{dx}\ \varphi(x) = \frac{x}{2}\ (\frac{\pi \cosh \pi x}{\sinh \pi x} - \frac{1}{x}) =  \frac{1}{2}\ (\pi x \coth \pi x -1) (4)

    Kind regards

    \chi \sigma
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