Lattices

Triple sum and summation theorem

Today we are going to prove this triple sum posted by @infesieres on Twitter. \[\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = \frac{ 7\pi^4}{720} + \frac{\pi^2\beta(2)}{6}\]
Proof.

First, recall the following super theorem of complex analysis which we will be using extensively:

Let $f$ be analytic in $\mathbb{C}$ except for finitely isolated singularities. If there is a constant $M>0$ such that $\displaystyle |f(z)|\leq \frac{M}{z^{k}}$ where $k>1$ then we have the summation formula \[ \lim_{N \to \infty } \sum_{n=-N}^{n=N} \{f(n) | \; n \textrm{ is not a singularity of } f\} = - \sum \{\textrm{Residues of } \pi\cot \pi z \;\textrm{ at the singularities of } f \}\] If none of the singularities of $f$ are at integers, then $\displaystyle \lim_{N \to \infty} \sum_{n=-N}^{n=N} f(n)$ exists, is finite and \[ \lim_{N \to \infty } \sum_{n=-N}^{n=N} f(n) = - \sum\{\textrm{Residues of } \pi\cot \pi z \; \textrm{ at the singularities of } f \}\]

Lets go back to the left hand side of the triple sum \[\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)} \sum_{c=1}^{\infty} \frac{1}{(a^2+c^2)} \] Consider the function $\displaystyle f(z) = \frac{1}{a^2+z^2}$. This function has two singularities at $z =\pm ia$. Moreover, $\displaystyle \lim_{z\to \infty} |z^2f(z)|=0$ so the conditions of the summation theorem hold. Then \begin{align*} \sum_{n=-\infty}^{\infty}\frac{1}{a^2+n^2} =& -\operatorname{Re}\left(\frac{\pi\cot \pi z}{a^2+z^2}, ia\right)- \operatorname{Re}\left(\frac{\pi\cot \pi z}{a^2+z^2}, -ia\right)\\ =& -\lim_{z\to ia} (z-ia)\frac{\pi\cot \pi z}{a^2+z^2} -\lim_{z\to -ia} (z+ia)\frac{\pi\cot \pi z}{a^2+z^2}\\ =& \frac{\pi\coth(a\pi)}{2a} +\frac{\pi\coth(a\pi)}{2a}\\ =&\frac{\pi \coth(a\pi)}{a} \end{align*} Now, we know that the function $f(n)$ is even, then: \begin{align*} \sum_{n=-\infty}^{\infty}\frac{1}{a^2+n^2} =&\sum_{n=-\infty}^{n=-1}\frac{1}{a^2+n^2} + \frac{1}{a^2}+ \sum_{n=1}^{\infty}\frac{1}{a^2+n^2} \\ =& \frac{1}{a^2} + 2 \sum_{n=1}^{\infty}\frac{1}{a^2+n^2}\\ \Longrightarrow \sum_{n=1}^{\infty}\frac{1}{a^2+n^2} =& \frac{1}{2}\sum_{n=-\infty}^{n=\infty}\frac{1}{a^2+n^2} -\frac{1}{2a^2}\\ =& \frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \end{align*} Then \begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} =& \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)} \sum_{c=1}^{\infty} \frac{1}{(a^2+c^2)}\\ =& \sum_{a=1}^{\infty}\left[\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)}\left(\frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \right)\right]\\ =& \sum_{a=1}^{\infty}\left[\left(\frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \right)\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)}\right]\\ =& \sum_{a=1}^{\infty}\left(\frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \right)^2\\ =& \sum_{a=1}^{\infty}\left[\frac{1}{4a^4}-\frac{\pi \coth \pi a}{2a^{3}}+\frac{\pi^2\coth^2\pi a}{4a^2} \right]\\ =& \frac{1}{4} \sum_{a=1}^{\infty} \frac{1}{a^4} - \frac{1}{2}\sum_{a=1}^{\infty}\frac{\pi \coth \pi a}{a^{3}} + \frac{1}{4}\sum_{a=1}^{\infty}\frac{\pi^2\cot^2\pi a}{a^2}\\ =& \frac{\pi^4}{360} - \frac{1}{2}\underbrace{\sum_{a=1}^{\infty}\frac{\pi \coth \pi a}{a^{3}}}_{A} + \frac{1}{4}\underbrace{\sum_{a=1}^{\infty}\frac{\pi^2\coth^2\pi a}{a^2}}_{B} \end{align*} We just have to find $A$ and $B$. Lets start with $A$ which is a relatively well known result of complex analysis. Consider the function $k(z)=\displaystyle \frac{\pi \coth \pi z}{z^{3}}$, it has a singularity at $z=0$ and countable set of singularities at $z=\pm in \quad n\in \mathbb{N}$. Additionally, $\displaystyle \lim_{z\to \infty} |z^2k(z)| =0$, so the assumptions of the summation theorem apply for the non-integer singularities. Then \begin{align*} \sum_{n=-\infty }^{\infty } \frac{\pi \coth \pi n}{n^{3}} = &-\sum_{n=1}^{\infty}\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},in\right)- \sum_{n=1}^{\infty}\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},-in\right)-\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right)\\ =& -\sum_{n=1}^{\infty}\left(\lim_{z\to in} (z-in)\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}}\right)-\sum_{n=1}^{\infty}\left(\lim_{z\to -in} (z+in)\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}}\right)-\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right)\\ =& -\sum_{n=0}^{\infty} \frac{\pi \coth \pi n}{n^3}-\sum_{n=0}^{\infty} \frac{\pi \coth \pi n}{n^3}-\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right) \end{align*} For $\displaystyle \operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right)$ we cannot apply the summation theorem directly (given that the singularity is an integer) but we can expand the function $k(z)$ with the Laurent series: \begin{align*} \frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}} =& \frac{\pi^2}{z^3} \left(\frac{1}{\pi z} -\frac{\pi z}{3} -\frac{\pi^3z^3}{45}+\right)\left(\frac{1}{\pi z} +\frac{\pi z}{3} -\frac{\pi^3z^3}{45}+\right)\\ =& \frac{1}{z^5}-\frac{7\pi^4}{45z} - \frac{19 \pi^9 z^3}{14175} -... \end{align*} Therefore $\displaystyle \operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right) =-\frac{7\pi^4}{45} $ Given that $k(z)$ is even \[4 \sum_{n=1}^{\infty } \frac{\pi \coth \pi n}{n^{3}} = \frac{7\pi^4}{45} \Longrightarrow \sum_{n=1}^{\infty } \frac{\pi \coth \pi n}{n^{3}} = \frac{7\pi^4}{180}=A\] Now for B first recall the following identity: $\coth^2 x = \operatorname{csch}^2x+1$. Therefore \[B= \sum_{n=1}^{\infty} \frac{\pi^2\coth^2 \pi n}{n^2} = \underbrace{\sum_{n=1}^{\infty} \frac{\pi^2\operatorname{csch}^2 \pi n}{n^2}}_{C} + \frac{\pi^4}{6}\] It turned out that we found C before in this post (actually the triple sum and that result seems to be deeply related): \[C = \sum_{n=1}^{\infty} \frac{\pi^2 \operatorname{csch}^2 \pi n}{n^2 } = \frac{2\pi^2}{3}\beta(2)-\frac{11}{180}\pi^4\] Therefore \[B =\frac{2\pi^2}{3}\beta(2)-\frac{11}{180}\pi^4 + \frac{\pi^4}{6}\] Then \begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = & \frac{\pi^4}{360} - \frac{1}{2}\underbrace{\sum_{a=1}^{\infty}\frac{\pi \coth \pi a}{a^{3}}}_{A} + \frac{1}{4}\underbrace{\sum_{a=1}^{\infty}\frac{\pi^2\coth^2\pi a}{a^2}}_{B}\\ =& \frac{\pi^4}{360} - \frac{1}{2} \underbrace{\frac{7\pi^4}{180}}_{A} + \frac{1}{4} \underbrace{\left[\frac{2\pi^2}{3}\beta(2)-\frac{11}{180}\pi^4 +\frac{\pi^4}{6}\right]}_{B} \\ =& \frac{ 7\pi^4}{720} + \frac{\pi^2\beta(2)}{6} \end{align*} Therefore \[\boxed{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = \frac{ 7\pi^4}{720} + \frac{\pi^2\beta(2)}{6}}\]