The funcation series arise in the invastigation of a interesting problem. How to find the sum? help in need. Here it is:

$\displaystyle \sum \frac{x^2}{k^2+x^2}$

the sum is taken over all positive integers $\displaystyle k$.

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- Mar 9th 2011, 01:38 AMShanksThe 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:

$\displaystyle \sum \frac{x^2}{k^2+x^2}$

the sum is taken over all positive integers $\displaystyle k$. - Mar 9th 2011, 03:47 AMFernandoRevilla
You can use the following result:

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

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

You'll obtain $\displaystyle S=\pi x \coth (\pi x)$ where $\displaystyle x>0$. - Mar 9th 2011, 07:48 AMShanks
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! - Mar 9th 2011, 08:08 AMchisigma
Alternatively You can start from the [well known...] 'infinite product'...

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

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

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

... and then...

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

... so that is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$