Sum of $\sum_0^\infty \frac1{n^4+a^4}$ How to find the following series as pretty closed form by $a$?
$$S=\sum_0^\infty \frac1{n^4+a^4}$$
I first considered applying Herglotz trick, simply because the expressions are similar. So I changed it like this...
$$2S-a^{-4}=\sum_{-\infty}^\infty \frac1{n^4+a^4}$$
However, the attempt failed to find such an appropriate function like $\pi\cot\pi z$ in this post.
Next I found this post and used Fourier transform in a similar way, and the result was a nightmare!

How on earth can I calculate the value of this series?

We start with partial fractions,
$$\frac{1}{n^4+a^4} = \frac{i}{2a^2}\left(\frac{1}{n^2 + i a^2} - \frac{1}{n^2 - i a^2}\right) \tag 1$$
And then note that (Series expansion of $\coth x$ using the Fourier transform)
$$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2}$$
which we rearrange to give:
$$\sum_{n=1}^\infty \frac{1}{\left(\frac{x}{\pi}\right)^2+n^2} = \frac{\pi^2}{2x}\left(\coth x - \frac{1}{x} \right) $$
$$\sum_{n=0}^\infty \frac{1}{\left(\frac{x}{\pi}\right)^2+n^2} = \frac{\pi^2}{x^2}+\frac{\pi^2}{2x}\left(\coth x - \frac{1}{x} \right) =\frac{\pi^2}{2x^2}\left(x \coth x + 1 \right) \tag 2$$
and so
$$\sum_{n=0}^\infty \frac{1}{n^4+a^4} = \frac{i}{2a^2}\sum_{n=0}^\infty \left(\frac{1}{n^2 + i a^2} - \frac{1}{n^2 - i a^2}\right) $$
$$ = \frac{i}{2a^2}\frac{\pi^2}{2ia^2\pi^2}\left(\sqrt{i}a \pi \coth \sqrt{i}a - \sqrt{-i}a \pi \coth \sqrt{-i}a\right) $$
$$ = \frac{\pi}{4a^3}\left(e^{i\pi/4}  \coth \sqrt{i}a - e^{3i\pi/4} \coth \sqrt{-i}a\right) $$
which I'm sure can be further simplified but I will leave there for now.
