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When I do the sum

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

the result reads

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

What is the meaning of True in the second result?

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

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

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closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
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See Piecewise. It is like the cases environment in $\LaTeX$. In native math, one would use "else" instead of True.

By the way, version 12 returns

$$\begin{cases} \frac{\coth \left(\sqrt{a}-i b\right)+\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}} & \arg \left(b+i \sqrt{a}\right)\geq 0 \\ \frac{\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}}+\frac{\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}}-\frac{1}{\sqrt{a}} & \text{True} \end{cases}$$

to be interpreted as

$$\begin{cases} \frac{\coth \left(\sqrt{a}-i b\right)+\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}} & \arg \left(b+i \sqrt{a}\right)\geq 0 \\ \frac{\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}}+\frac{\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}}-\frac{1}{\sqrt{a}} & \text{else} \end{cases}$$

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  • 1
    $\begingroup$ Ok, so it really simply means "otherwise" or "else" or "in all other cases"? $\endgroup$ – wondering May 22 at 18:44
  • $\begingroup$ Yes. That is correct. $\endgroup$ – Henrik Schumacher May 22 at 18:45

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