Is this series known? [closed]

I encountered the following series and was wondering if it can be expressed using a known value?

$$\sum_{a,b=1}^{\infty} \frac{(-1)^a+1}{a^3-a^2+ab^2+b^2},$$ where all terms in the sum for which $a$ is odd are interpreted as zero.

Solution 1:

Since $a$ can only contribute nontrivially when even, we may as well replace $a\to 2a$

$$\sum_{(a,b)\in\Bbb{Z}^+\times\Bbb{Z}^+} \frac{2}{8a^3-4a^2+(2a+1)b^2} = \sum_{a=1}^\infty \frac{2}{2a+1}\sum_{b=1}^\infty\frac{1}{4a^2\left(\frac{2a-1}{2a+1}\right)+b^2}$$

The sum on the interior is the famous result

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

leaving us with

$$\sum_{a=1}^\infty \frac{2}{2a+1}\left[\frac{\pi}{4a}\sqrt{\frac{2a+1}{2a-1}}\coth\left(2\pi a\sqrt{\frac{2a-1}{2a+1}}\right) - \frac{1}{8a^2}\left(\frac{2a+1}{2a-1}\right)\right]$$

$$ = \sum_{a=1}^\infty \frac{\pi}{2a}\frac{1}{\sqrt{4a^2-1}}\coth\left(2\pi a\sqrt{\frac{2a-1}{2a+1}}\right) - \frac{1}{4a^2}\frac{1}{2a-1}$$

Both terms converge by comparison test so the series converges, and the term on the right has an easy closed form

$$\sum_{a=1}^\infty \frac{-1}{4a^2(2a-1)} = \frac{1}{4}\sum_{a=1}^\infty\frac{1}{a^2} + \sum_{a=1}^\infty \frac{1}{2a}-\frac{1}{2a-1} = \frac{\pi^2}{24} - \log 2$$

However the term on the left with the $\coth $ would be tricky, and it is unlikely a nice closed form would exist.