<s>[INST] Can we give a closed form expression for $\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}$? Let $a>0$ and $x\in(-1,1)$. Can we give a closed form expression for $\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}$? Note that the series is convergent. There should be an expression in terms of the cotangent.
My goal is to find a (sharp) upper bound $c(x)$ of the series.
 [/INST] Observe
\begin{align}
\sum^\infty_{k=-\infty}\frac{1}{a+(x+k)^2} =&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\
&+\sum^\infty_{k=1} \frac{1}{a+(k+(1+x))^2}+\sum^{\infty}_{k=1} \frac{1}{a+(k+(1-x))^2}\\
\leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\
&\ +\int^\infty_0\frac{dk}{a+(k+(1+x))^2}+ \int^\infty_{0} \frac{dk}{a+(k+(1-x))^2}\\
\leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\
&\ +\frac{1}{\sqrt{a}}\left(\tan^{-1}\left( \frac{\sqrt{a}}{1+x}\right)+\tan^{-1}\left( \frac{\sqrt{a}}{1-x}\right)\right)
\end{align}
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