<s>[INST] Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ How to show the following equality?
$$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$
 [/INST] Related problems: (I), (II). This problem is a direct application of Fourier transform and Poisson summation formula. Recalling the definition of Fourier transform and the Poisson summation formula respectively

$$ F(w) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-ixw} dx \,,$$
$$ \sum_{-\infty}^{\infty} f(n) = \sqrt{2\pi}\sum_{-\infty}^{\infty} F(2n\pi)\,, $$

where $F$ is the Fourier transform of $f$. Advancing with our problem, first, we compute the Fourier transform of $ f(x)=\frac{1}{x^2+a^2} $ which is equal to

$$ F(w) = \sqrt{\frac{\pi}{2}}\frac{1}{a}e^{-a|w|}\,.$$

Applying Poisson formula, we have
$$ \sum_{n=0}^{\infty}\frac{1}{n^2+a^2} = \frac{\pi}{a}\sum_{n=0}^{\infty}e^{-2an\pi} = \frac{\pi}{a} \sum_{n=0}^{\infty}r^{n}=\frac{\pi}{a}\frac{1}{1-r}\,,\quad r = e^{-2 \pi a} \,,$$
$$\Rightarrow \sum_{n=0}^{\infty}\frac{1}{n^2+a^2} = \frac{\pi}{a} \frac{1}{1-e^{-2a\pi}}=\frac{\pi}{a} \frac{e^{2a\pi}}{e^{2a\pi}-1} \,. $$
Now, I leave it to you to manipulate the above expression to reach the form

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

You can use the identity

$$ \coth x = \frac{\cosh x}{\sinh x} = \frac {e^x + e^{-x}} {e^x - e^{-x}} = \frac{e^{2x} + 1} {e^{2x} - 1} \,. $$

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