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?

If you write
$$n^4+a^4=(n-\alpha)(n-\beta)(n-\gamma)(n-\delta)$$
$$\alpha=-\frac{(1+i) a}{\sqrt{2}}\qquad \beta=\frac{(1+i) a}{\sqrt{2}}\qquad \gamma=-\frac{(1-i) a}{\sqrt{2}} \qquad \delta=\frac{(1-i) a}{\sqrt{2}}$$ Using partial fraction
$$\frac 1{n^4+a^4}=\frac{1}{(\alpha-\beta) (\alpha-\gamma) (\alpha-\delta) (x-\alpha)}+\frac{1}{(\beta-\alpha) (\beta-\gamma) (\beta-\delta) (x-\beta)}+$$ $$\frac{1}{(\gamma-\alpha) (\gamma-\beta) (\gamma-\delta) (x-\gamma)}+\frac{1}{(\delta-\alpha) (\delta-\beta) (\delta-\gamma) (x-\delta)}$$
Now,consider the partial sum
$$S_p(\epsilon)=\sum_{n=0}^p \frac 1{n-\epsilon}=\psi (p+1-\epsilon )-\psi (-\epsilon )$$  Compute all sums and use the asymptotics of the digamma function for large $p$ to obtain for large $p$
$$S_p=\sum_{n=0}^p \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi  a\right)+\sin \left(\sqrt{2} \pi  a\right)}{\cosh
   \left(\sqrt{2} \pi  a\right)-\cos \left(\sqrt{2} \pi  a\right)}-\frac{1}{3 p^3}+O\left(\frac{1}{p^4}\right)$$
$$\color{red}{S_\infty=\sum_{n=0}^\infty \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi  a\right)+\sin \left(\sqrt{2} \pi  a\right)}{\cosh
   \left(\sqrt{2} \pi  a\right)-\cos \left(\sqrt{2} \pi  a\right)}}$$
