# 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.

• Assume $a\neq 0$ and use contour integration with the meromorphic function $z\mapsto\frac{\cot(\pi z)}{z^4+a^4}$ over the circle $C\left(0;n+\frac 12\right)$. – Philippe Malot May 7 '13 at 19:26
• I don't know the procedure, but at least a possible solution: Wolfram|Alpha can be good at these. – apnorton May 7 '13 at 19:27
• This is going to be heavily related to some linear combination of series expansions of $cot(z)$ and $coth(z)$. Wolfram Alpha confirms this and gives an explicit form: wolframalpha.com/input/… Check out the very last identity of series expansions of $cot$ here: mathworld.wolfram.com/Cotangent.html – Alex R. May 7 '13 at 19:28
• I just looked at the wolphram alpha solution. How in the... – Zainy May 7 '13 at 19:31
• Do you want a more detailed answer? – Philippe Malot May 7 '13 at 19:41

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.

• @girianshiido: thanks, I really appreciate it. The part that's terrifying about all of this is the algebra involved when the degree of the denominator is $> 2$. I hope I gave enough detail, yet not too much that I essentially did it all for the OP. – Ron Gordon May 7 '13 at 20:21
• Whoa! Dadgum, thank you man. The only experience I've ever had with the residue theorem was in the use of contour integration. I would have never thought of it. I'm going to sift through this for the next couple of hours or so, what would make for a good resource? Does MIT have any open courseware on the residue theorem? I'd be interested in learning it. ${EDIT}:$ Interested in learning the residue theorem solution for series expansions. – Zainy May 7 '13 at 22:14
• @Zainy: funny that. No real help on this via MIT OCW. I really got everything I needed from Ahlfors and a very very good professor (no coincidence that the three courses I took with him from 1989-91 are the ones in which I contribute the most in M.SE). In the meantime, here's a decent, short ref: faculty.up.edu/wootton/Complex/Chapter11.pdf – Ron Gordon May 7 '13 at 22:26
• @RonGordon : Nice document & Bonus. Indeed the algebra involved here is quite terrifying. The $\cos(b e^it)$ formula is helpful. – Philippe Malot May 8 '13 at 12:05
• Could this be generalized to $\dfrac1{z^k+a^k}$? – marty cohen Jul 27 '14 at 1:26

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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}}} = {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}}}} ={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}}} =-\,{\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}} \qquad\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}

$$\color{#00f}{\begin{array}{|l|}\hline\\ \quad{\rm J}\pars{a} =\sum_{n = -\infty}^{\infty}{1 \over n^{4} + a^{4}} \\[3mm]\quad ={\root{2}\pi \over 2}\, {\tan\pars{\root{2}\pi a/2}\sech^{2}\pars{\root{2}\pi a/2} +\tanh\pars{\root{2}\pi a/2}\sec^{2}\pars{\root{2}\pi a/2} \over a^{3}\bracks{\tan^{2}\pars{\root{2}\pi a/2} + \tanh^{2}\pars{\root{2}\pi a/2}}} \quad \\ \\ \hline \end{array}}$$ We can check, after a "painful manipulation", that the above expression satisfies $\ds{\lim_{a \to 0} \underbrace{{1 \over 2}\bracks{{\rm J}\pars{a} - {1 \over a^{4}}}} _{\ds{=\ \sum_{n = 1}^{\infty}{1 \over n^{4}}}} = {\pi^{4} \over 90}}$.

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 $0\notin Q(\Bbb R)$

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 $(3)$ : $$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh\left(\sqrt{2} \pi a\right)+\sin\left(\sqrt{2} \pi a\right)}{\cosh\left(\sqrt{2} \pi a\right) -\cos\left(\sqrt{2} \pi a\right)}$$