The sum of function series

**Shanks** · March 9th 2011, 01:38 AM

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 $k$ .

**FernandoRevilla** · March 9th 2011, 03:47 AM

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 $x>0$ .

**Shanks** · March 9th 2011, 07:48 AM

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!

**chisigma** · March 9th 2011, 08:08 AM

Originally Posted by Shanks

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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