A double lattice sum

Let $s \in \mathbb{C}$ such that $\mathfrak{Re}(s)>1$ . Evaluate the following double Euler sum

$\mathcal{S}= \sum_{\left ( j, k \right ) \in \mathbb{Z}^2 \setminus \left \{ (0, 0) \right \}} \frac{1}{\left ( j^2+k^2 \right )^s}$

Solution

The number of ways of writing an integer as sum of two square integers (both positive and negative) is known to be

$r_2(n) = 4\sum\limits_{\substack{d |n\\ d \text{ odd }}} (-1)^{\frac{d-1}{2}}$

Then for $\mathfrak{Re}(s)>1$ we have that

$\begin{align*} \sum_{(j,k) \in \mathbb{Z}^2 \setminus \{(0,0)\}} \frac{1}{(j^2+k^2)^s} &= \sum_{n=1}^{\infty} \frac{r_2(n)}{n^s} \\ &= 4\sum_{\substack{d \ge 1\\d \text{ odd }}} \sum\limits_{n \equiv 0 (d)} \frac{(-1)^{\frac{d-1}{2}}}{n^s} \\ &= 4\sum\limits_{\substack{d \ge 1\\d \text{ odd }}} \sum\limits_{m=1}^{\infty} \frac{(-1)^{\frac{d-1}{2}}}{m^sd^s} \\ &= 4\sum\limits_{d \ge 1} \frac{(-1)^{d-1}}{(2d-1)^s}\sum\limits_{m=1}^{\infty} \frac{1}{m^s} \\ &= 4\beta(s)\zeta(s) \end{align*}$

where $\beta$ is the Beta Dirichlet function and $\zeta$ is the Riemann zeta function.

This particular solution may be found at mathimatikoi.org forum and of course the interested reader will find some nice applications of that sum.

A double lattice sum

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