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A double lattice sum

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

\displaystyle \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

\displaystyle 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{aligned} \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{aligned}

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.

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