<s>[INST] Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ I've been working with series this week and came across a couple that have been bugging me.
I'm looking for the closed form of:
$$
J(a)=\sum_{-\infty}^{\infty} \frac {1} {n^4+a^4}
$$
As with the other one I posted, solving for the closed form seems like a futile effort as nothing involving differentiation, integration, or DE form leads to a known series solution, such as $e^{x}$, $ln(x)$, etc.  I'm running out of hair here.  I'll thank anyone who's willing to help.
 [/INST] Rather "magical" approach :
Recall the Mittag-Leffler meromorphic expansion of the $\cot$ ($\coth$, resp.) function :
$$\sum_{n=-\infty}^{\infty}\frac{1}{n^2-a^2} = -\frac{\pi\cot\pi a}{a} $$
holding for any complex $a$ (exept the poles obviously) - this fact we will use later as change $a\rightarrow a i$.
From this we have
$$\sum_{n=-\infty}^{\infty}\frac{1}{(n\!-\!a)^2\!-\!b^2} = \frac{1}{2b}\sum_{n=-\infty}^{\infty}\frac{a-b}{n\!-\!(a\!-\!b)^2}\!+\!\frac{a+b}{n\!-\!(a\!+\!b)^2} =\frac{\pi}{2b}\left[\cot\pi (a\!-\!b)\!-\!\cot\pi (a\!+\!b)\right]\tag{1}$$
Similarly we can show
$$\sum_{n=-\infty}^{\infty}\frac{n+a}{(n\!+\!a)^2\!-\!b^2}\!-\!\frac{n-a}{(n\!-\!a)^2\!-\!b^2} = \pi\left[\cot\pi (a\!+\!b)\!+\!\cot\pi (a\!-\!b)\right]\tag{2}$$
Start with
$$n^4\!+\!a^4\!=\!n^4\!+\!2a^2n^2\!+\!a^4\!-\!2a^2n^2\!=\!\left(n^2\!+\!a^2\right)^2\!-\!\left(\sqrt{2}an\right)^2\!=\!\left(n^2\!+\!na\sqrt{2}\!+\!a^2\right)\left(n^2\!-\!na\sqrt{2}\!+\!a^2\right)$$
Then, using partial fraction expansion :
$$\begin{align*}&\sum_{n=-\infty}^{\infty}\frac{1}{n^4+a^4}=\frac{1}{2\sqrt{2}a^3}\sum_{n=-\infty}^{\infty}\frac{n+\sqrt{2}a}{n^2+\sqrt{2}an+a^2}-\frac{n-\sqrt{2}a}{n^2-\sqrt{2}an+a^2} = \\ \\
&\frac{1}{2\sqrt{2}a^3}\sum_{n=-\infty}^{\infty}\frac{n+\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}}{\left(n+\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{\sqrt{2}}\right)^2}-\frac{n-\frac{a}{\sqrt{2}}-\frac{a}{\sqrt{2}}}{\left(n-\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{\sqrt{2}}\right)^2}\end{align*} $$
By (1) and (2) we immidiatelly conclude this equals :
$$\frac{\pi}{2\sqrt{2}a^3}\left[\cot \left(\pi a\frac{1+i}{\sqrt{2}}\right)\!+\!\cot \left(\pi a\frac{1-i}{\sqrt{2}}\right)\!-i\cot \left(\pi a\frac{1-i}{\sqrt{2}}\right)\!+i\cot \left(\pi a\frac{1+i}{\sqrt{2}}\right)\right] $$
Denote $\alpha := \pi a(1-i)/\sqrt{2}$, then $\bar{\alpha}=\pi a(1+i)/\sqrt{2}$, and since $\cot\bar{\alpha}=\overline{\cot\alpha}$ meromorphic :
$$\sum_{n=-\infty}^{\infty}\frac{1}{n^4+a^4}=\frac{\pi}{2\sqrt{2}a^3}\left[\cot\bar{\alpha} + \cot\alpha -i\cot\alpha+i\cot\bar{\alpha}\right]=\frac{\pi}{\sqrt{2}a^3}\left[\Re\cot \alpha + \Im\cot \alpha\right]\tag{3} $$
Now a bit of half angle trigonometry, from wiki for example :
$$\cot\left(\frac{A-B}{2}\right)=\frac{\sin A-\sin B}{\cos B - \cos A}$$
This implies, since $\sin ix=i \sinh x$ and $\cos ix = \cosh x$ :
$$\cot \alpha=\frac{\sin\left(\sqrt{2} \pi a\right)+i\sinh\left(\sqrt{2} \pi a\right)}{\cosh\left(\sqrt{2} \pi a\right) -\cos\left(\sqrt{2} \pi a\right)}$$
And therefore by $(3)$ :
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh\left(\sqrt{2} \pi a\right)+\sin\left(\sqrt{2} \pi a\right)}{\cosh\left(\sqrt{2} \pi a\right) -\cos\left(\sqrt{2} \pi a\right)}$$
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