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Show that $$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth\pi a+\frac{1}{2a^2}, a>0$$

I know I must use summation theorem and I calculated the residue which is:

$$Res\left(\frac{1}{z^2+a^2}, \pm ai\right)=-\frac{\pi}{2a}\coth\pi a$$

Now my question is: how do I get the last term $+\frac{1}{2a^2}$ after using the summation theorem?

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The method of residues applies to sums of the form

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{res}_{z=z_k} \pi \cot{\pi z}\, f(z)$$

where $z_k$ are poles of $f$ that are not integers. So when $f$ is even in $n$, you may express as follows:

$$2 \sum_{n=1}^{\infty} f(n) + f(0)$$

For this case, $f(z)=1/(z^2+a^2)$ and the poles $z_{\pm}=\pm i a$ and using the fact that $\sin{i a} = i \sinh{a}$, we get

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \text{coth}{\pi a}$$

The rest is just a little more algebra.

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  • $\begingroup$ Thanks, Ron. As for the sum in question, do we just take half of this whole expression? $\endgroup$ – User69127 May 20 '13 at 0:34
  • $\begingroup$ @User69127: not quite. See the above expression. $\endgroup$ – Ron Gordon May 20 '13 at 0:37
  • $\begingroup$ I guess it's the algebra that has me confused. So taking half of this sum, I obtain the $\frac{\pi}{2a}$ portion and the $\frac{1}{2a^2}$ since i'm only interesting in the n=0 to infinity portion, right? $\endgroup$ – User69127 May 20 '13 at 0:46
  • $\begingroup$ Let the sum from 1 to infinity be $S$ and the sum from $-\infty$ to $\infty$ be $RHS$. Then $2 S+f(0)=RHS \implies S=\frac12 (RHS - f(0))$ and therefore the sum from 0 to infinity = $S + f(0) = \frac12 (RHS+f(0))$. $\endgroup$ – Ron Gordon May 20 '13 at 1:03
  • $\begingroup$ Oh I see - that's exactly what I meant but I must of said it wrong lol. Thank you!! $\endgroup$ – User69127 May 20 '13 at 1:08

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