TITLE: An interesting sum over lattice points in a large disk centered at the origin
QUESTION [10 upvotes]: Evaluate the the limit, as $r \rightarrow \infty $, of the sum $\displaystyle \sum \limits_{(m,n) \in D_r}$ $\displaystyle (-1)^{m+n} \over \displaystyle m^2 + n^2$, where $D_r$ denotes the closed disk of radius $r$ centered at the origin.  
I expect one would need Poisson summation to turn this into an exponential sum and apply some analytic estimate, but I cannot find any relevant results. Any help and especially, references, would be welcome.

REPLY [11 votes]: It is problem number 10 of IMC 2018, you may find the solution on the official site.

REPLY [9 votes]: More generally, when $\Re(s)\ge1$
$$\sum_{(m,n)\ne(0,0)}(-1)^{m+n}/(m^2+n^2)^s=-4(1-2^{1-s})\zeta_K(s)$$
where $K=\mathbb Q(i)$ and $\zeta_K$ is the Dedekind zeta function of $K$.

REPLY [4 votes]: Assuming convergence, here goes:
let $r_2(k)$ denote the number of ways a number $k\in\mathbb{N}$ can be written
as the sum of squares of two integers. Then, we compute
$$\sum_{(0,0)\neq(m,n)\in\mathbb{Z}^2}\frac{(-1)^{m+n}}{m^2+n^2}=\sum_{k=1}^{\infty}(-1)^k\frac{r_2(k)}k.$$
It is known that
$$r_2(k)=4\sum_{d\vert k}\left(\frac{-4}k\right)=4\left(1*\left(\frac{-4}k\right)\right)(k)$$
where $\left(\frac{a}b\right)$ is the Jacobi symbol. Using $\left(\frac{-4}{2k}\right)=0$ and $\left(\frac{-4}{2k+1}\right)=(-1)^k$ it follows that
\begin{align} \sum_{k=1}^{\infty}(-1)^k\frac{r_2(k)}k
&=4\sum_{k=1}^{\infty}(-1)^k\frac{(1*\left(\frac{-4}k\right)(k)}k \\
&=4\sum_{n=1}^{\infty}\frac{(-1)^n}n\sum_{k=1}^{\infty}(-1)^k\frac{\left(\frac{-4}k\right)(k)}k \\
&=4\sum_{n=1}^{\infty}\frac{(-1)^n}n\sum_{m=0}^{\infty}\frac{(-1)^m}{2m+1} \\
&=4(-\log(2))\left(\frac{\pi}4\right) \\
&=-\pi\,\log(2),
\end{align}
which confrims Henri Cohen's guesstimate.