# What is the meaning of True in my result? [closed]

When I do the sum

Sum[(a + (b + π n)^2)^(-1), {n, -∞, ∞}]


$$\begin{array}{cc} \{ & \begin{array}{cc} \frac{\coth \left(\sqrt{a}+i b\right)+\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}} & \arg \left(\sqrt{a}-i b\right)\geq -\frac{\pi }{2}\land \arg \left(\sqrt{a}-i b\right)\leq \frac{\pi }{2} \\ \frac{\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}}+\frac{\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}}-\frac{1}{\sqrt{a}} & \text{True} \\ \end{array} \\ \end{array}$$

What is the meaning of True in the second result?

If I get rid of $$\pi$$, the result is simply:

$$\frac{-\pi \cot \left(\pi \sqrt{-a}+\pi b\right)-\pi \cot \left(\pi \sqrt{-a}-\pi b\right)}{2 \sqrt{-a}}$$

See Piecewise. It is like the cases environment in $$\LaTeX$$. In native math, one would use "else" instead of True.
$$\begin{cases} \frac{\coth \left(\sqrt{a}-i b\right)+\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}} & \arg \left(b+i \sqrt{a}\right)\geq 0 \\ \frac{\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}}+\frac{\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}}-\frac{1}{\sqrt{a}} & \text{True} \end{cases}$$
$$\begin{cases} \frac{\coth \left(\sqrt{a}-i b\right)+\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}} & \arg \left(b+i \sqrt{a}\right)\geq 0 \\ \frac{\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}}+\frac{\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}}-\frac{1}{\sqrt{a}} & \text{else} \end{cases}$$