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.
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}
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 \newcommand{\ic}{{\rm i}}
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$\ds{
{\rm J}\pars{a} = \sum_{n = -\infty}^{\infty}{1 \over n^{4} + a^{4}}
=
-\,{1 \over a^{4}} + 2\sum_{n = 0}^{\infty}{1 \over n^{4} + a^{4}}
}$

\begin{align}
&{\rm J}\pars{a} + {1 \over a^{4}}
=
2\sum_{n = 0}^{\infty}{1 \over 2\ic a^{2}}
\pars{{1 \over n^{2} - \ic a^{2}} - {1 \over n^{2} + \ic a^{2}}}
=
{2 \over a^{2}}\Im\sum_{n = 0}^{\infty}{1 \over n^{2} - \ic a^{2}}
\\[3mm]&=
{2 \over a^{2}}\Im\sum_{n = 0}^{\infty}
{1 \over \pars{n + \tilde{n}}\pars{n - \tilde{n}}}
\qquad\mbox{where}\qquad \tilde{n} = \expo{\ic\pi/4}\verts{a}
\end{align}

\begin{align}
&{\rm J}\pars{a} + {1 \over a^{4}}
=
{2 \over a^{2}}\Im\bracks{
\Psi\pars{\tilde{n}} - \Psi\pars{-\tilde{n}} \over \tilde{n} - \pars{-\tilde{n}}}
\\[3mm] = &\
{1 \over \verts{a}^{3}}\Im\braces{\expo{-\ic\pi/4}\bracks{
\Psi\pars{\tilde{n}} - \Psi\pars{-\tilde{n}}}}\tag{1}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function
$\bf\mbox{6.3.1}$.

Also,
$$
\Psi\pars{\tilde{n}} - \Psi\pars{-\tilde{n}}
=\overbrace{\Psi\pars{\tilde{n} + 1} - {1 \over \tilde{n}}}
^{\ds{=\ \Psi\pars{\tilde{n}}}}\ -\
\Psi\pars{-\tilde{n}}
=-\,{1 \over \tilde{n}} +\
\overbrace{\bracks{\Psi\pars{\tilde{n} + 1} - \Psi\pars{-\tilde{n}}}}
^{\ds{=\ \pi\cot\pars{\pi\bracks{-\tilde{n}}}}}
$$
where we used the Recurrence Formula
$\ds{\bf\mbox{6.3.5}}$ and the Euler Reflection Formula $\ds{\bf\mbox{6.3.7}}$. We replace this result in expression $\pars{1}$:
\begin{align}
{\rm J}\pars{a} + {1 \over a^{4}}&
={1 \over \verts{a}^{3}}\,\Im\braces{\expo{-\pi\ic/4}\bracks{-\,{1 \over \expo{\pi\ic/4}\verts{a}} - \pi\cot\pars{\pi\tilde{n}}}}
\\[3mm] & ={1 \over \verts{a}^{4}} - {\pi \over \verts{a}^{3}}
\,\Im\bracks{\expo{-\pi\ic/4}\cot\pars{\pi\tilde{n}}}
\end{align}

\begin{align}{\rm J}\pars{a} & = -\,{\pi \over \verts{a}^{3}}
\,\Im\bracks{\expo{-\pi\ic/4}\cot\pars{\pi\tilde{n}}}
\\[3mm] & =
-\,{\pi \over a^{3}}\,\Im\bracks{\expo{-\pi\ic/4}
\cot\pars{{\pi\root{2} \over 2}\, a + {\pi\root{2} \over 2}\, a\ic}}
\\[3mm]&=-\,{\root{2}\pi \over 2a^{3}}\,\Im\bracks{\pars{1 - \ic}\,
{1 - \mu\nu\ic \over \mu + \nu\ic}}
\\[3mm] & \mbox{where}\qquad\left\lbrace\begin{array}{rcl}
\mu & \equiv & \tan\pars{{\root{2}\pi \over 2}\,a}
\\[2mm]
\nu & \equiv & \tanh\pars{{\root{2}\pi \over 2}\,a}
\end{array}\right.
\\[5mm]
{\rm J}\pars{a}&={\root{2}\pi \over 2}\,
{\mu\pars{1 - \nu^{2}} + \nu\pars{1 + \mu^{2}} \over a^{3}\pars{\mu^{2} + \nu^{2}}}
\end{align}

$$\color{#00f}{\begin{array}{|l|}\hline\\
\ds{\quad{\rm J}\pars{a}
=\sum_{n = -\infty}^{\infty}{1 \over n^{4} + a^{4}}\quad}
\\[3mm]\ds{\quad
={\root{2}\pi \over 2}\,
{\tan\pars{\root{2}\pi a/2}\sech^{2}\pars{\root{2}\pi a/2}
+\tanh\pars{\root{2}\pi a/2}\sec^{2}\pars{\root{2}\pi a/2}
\over
a^{3}\bracks{\tan^{2}\pars{\root{2}\pi a/2} + \tanh^{2}\pars{\root{2}\pi a/2}}}
\quad\phantom{AAAAAAA}}
\\ \hline
\end{array}}
$$
We can check, after a "painful manipulation", that the above expression satisfies
$\ds{\lim_{a \to 0}
\underbrace{{1 \over 2}\bracks{{\rm J}\pars{a} - {1 \over a^{4}}}}
_{\ds{=\ \sum_{n = 1}^{\infty}{1 \over n^{4}}}} = {\pi^{4} \over 90}}$.
A: 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 $(3)$ :
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh\left(\sqrt{2} \pi a\right)+\sin\left(\sqrt{2} \pi a\right)}{\cosh\left(\sqrt{2} \pi a\right) -\cos\left(\sqrt{2} \pi a\right)}$$
A: As stated above, this sum may be evaluated using residue theory.  I will state the result: for $f$ sufficiently "well-behaved" (meaning that it vanishes sufficiently fast  along the vertical sections of the typical rectangular contour used to derive the following relation):
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \operatorname*{Res}_{z=z_k} [\pi \, \cot{\pi z} \, f(z)]$$
where $z_k$ is a non-(real integral) pole of $f$ in the complex plane.
In this case, $f(z)=1/(z^4+a^4)$ and the poles are at $z = a\,e^{i (2 k-1) \pi/4}$, $k \in \{1,2,3,4\}$.  So, evaluation of the sum reduces to summing the residues at these poles:
$$-\sum_k \text{Res}_{z=z_k} \pi \, \cot{\pi z} \, f(z) = -\sum_{k=1}^4 \frac{\pi \cot{(\pi e^{i (2 k-1) \pi/4})}}{4 a^3 e^{i 3 (2 k-1) \pi/4}}  $$
Now, you may deduce that
$$\cot{(b\, e^{i t})} = \frac{\sin(2 b \cos{t})}{\cosh(2 b \cos{t})-\cos(2 b \sin{t})}-i \frac{ \sinh(2 b \sin{t})}{\cosh(2 b \cos{t})-\cos(2 b \sin{t})}$$
You may either verify this formula in a program like Wolfram Alpha or Mathematica, or you can derive this by using the cosine and sine addition theorems. 
The algebra involved can be potentially tedious and error-prone.  I will outline here a few tips to get to the correct result.  Rewrite the sum over the residues as (negative sign included):
$$\frac{\pi}{4 a^3} \left [ e^{-i 3 \pi/4} \frac{-\sin{(\sqrt{2} \pi a)} + i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} + \\ e^{-i 9 \pi/4} \frac{\sin{(\sqrt{2} \pi a)} + i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} + \\ e^{-i 15 \pi/4} \frac{\sin{(\sqrt{2} \pi a)} - i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} + \\e^{-i 21 \pi/4} \frac{-\sin{(\sqrt{2} \pi a)} - i\sinh{(\sqrt{2} \pi a)} }{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} \right ]$$ 
It should be plain that the exponentials can be reduced, and you end up with two pairs of complex conjugates if everything is done correctly. At this point I will leave the details to the reader and state the final result:
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh{(\sqrt{2} \pi a)}+\sin{(\sqrt{2} \pi a)}}{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}}$$
BONUS
As a check, you can derive the well-known formula
$$\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$$
by considering the behavior of the above result in the limit as $a \to 0$.  Note that
$$\sum_{n=1}^{\infty} \frac{1}{n^4+a^4} = \frac12 \left [ \sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} - \frac{1}{a^4}\right]$$
Now Taylor expand the result far enough to get a nonvanishing result:
$$ \frac{\sinh{(\sqrt{2} \pi a)}+\sin{(\sqrt{2} \pi a)}}{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}} \sim \frac{2 \sqrt{2} \pi a \left ( 1+ \frac{\pi^4 a^4}{30} \right )}{2 \pi^2 a^2 \left ( 1+ \frac{\pi^4 a^4}{90} \right )} \sim \frac{\sqrt{2}}{\pi a} \left ( 1+ \frac{\pi^4 a^4}{45} \right)  $$
Putting this altogether, one easily obtains the desired sum.
A: Let $f:z\mapsto \dfrac{\pi\cot(\pi z)}{z^4+a^4}$, with $a\in\Bbb R^*$. 
It is a meromorphic function over $\Bbb C$, and its poles are $z_k=a e^{i\left(\dfrac \pi4 +k\dfrac\pi2\right)}$, $(k=0,\dots,3)$ and every integers. They're all simple poles and none of these poles is on the circle $C_n=C(0,n+\frac 12)$ (traveled once in the counterclockwise direction).
Let $n\in\Bbb Z$. You have : 
$$\mathrm{Res}(f,n)=\displaystyle\lim_{z\to n}\ (z-n)\dfrac{\pi\cot(\pi z)}{z^4+a^4}=\dfrac \pi{\pi\tan'(\pi n)(n^4+a^4)}=\dfrac 1{n^4+a^4}$$
By residue theorem, you have :
$$\dfrac 1{2i\pi}\int_{C_n} f(z)dz=\sum_{k=0}^3 \mathrm{Res}(f,z_k)+\sum_{k=-n}^n \dfrac 1{n^4+a^4}$$
You can prove (that's the tricky part) that the LHS tends to $0$ when $n$ tends to $\infty$,
So you get : $$\sum_{n=-\infty}^{+\infty} \dfrac 1{n^4+a^4}=-\pi\sum_{k=0}^3 \mathrm{Res}(f,z_k)$$
I'll let you finish !
You can use that $\pi\cot(\pi z)$ trick to evaluate series of the form $\displaystyle\sum_{n=-\infty}^{+\infty} \dfrac{P(n)}{Q(n)}$ with $P$ and $Q$ two polynomials s.t. $\deg(P)<\deg(Q)-1$ and $0\notin Q(\Bbb R)$
