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Given $|x|\leq 1$, has the series

$$\sum_{n=0}^{\infty}\frac{x^n}{(n-a)^2+b^2}$$

a closed form expression in simple functions?

It is known that for $x=1$ from Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. this sums up to

$$\frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)}$$

Also for $x=-1$ exists simple result

Can it be generalized as mentioned above?

EDIT:

Closed form above for $x=1$ relates to a modified series $$\sum_{n=-\infty}^{\infty}\frac{x^{|n|}}{(n-a)^2+b^2}$$

which looks more promising for candidate for close-term expression via Poisson's summation formula as @N74 suggests.

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  • $\begingroup$ Maybe you can reuse the Poisson summation formula solution using $x^n=e^{n \log x}$ $\endgroup$ – N74 Jul 20 '16 at 23:26
  • $\begingroup$ For $x=1$ the expression you gave is for $\sum_{n=-\infty}^\infty$. According to the question you linked, the result for $\sum_{n=0}^\infty$ is in terms of the digamma function. $\endgroup$ – stewbasic Jul 20 '16 at 23:29
  • $\begingroup$ I guess there is some hypergeometric around or/and incomplete Beta$\ldots$ $\endgroup$ – Felix Marin Jul 20 '16 at 23:31
  • $\begingroup$ @stewbasic actually yes that is indeed true, nevertheless for this purpose I should modify the sum to $\sum_{n=-\infty}^{\infty}\frac{x^{|n|}}{(x-a)^n+b^n}$ since I was interested in power decay, that would be more apropriate question, however, I admit there MUST be a solution in gamma/digamma (in case $n=0$), and Mathematica confirms me as Felix Marin suggested, there is a raw hypergeometric solution (even for $n=0$ case). $\endgroup$ – Machinato Jul 21 '16 at 0:09
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As Felix Marin commented, there is a solution in terms of the incomplete Beta function.

Starting with $$\frac{1}{(n-a)^2+b^2}=\frac i{2b}\left(\frac{1}{n-(a-i b)}-\frac{1}{n-(a+i b)}\right)$$

$$\sum_{n=0}^{\infty}\frac{x^n}{(n-a)^2+b^2}=\frac i{2b}\left(x^{(a-i b)} B_x(-(a-i b),0)-x^{(a+i b)} B_x(-(a+i b),0) \right)$$ What I suspect is that this could also write using Hurwitz-Lerch transcendent function since $$\sum_{n=0}^{\infty}\frac{x^n}{n+k}=\Phi (x,1,k)$$

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