IMC 2018 · Day 2 · P10

Define $E_R = \{ (a, b) \in \mathbb{Z}^2 \setminus \{(0,0)\} : a^2 + b^2 < R \text{ and } a + b \text{ is even} \}$. Then $$\tag{5} \sum_{(a,b) \in D_R} \frac{(-1)^{a+b}}{a^2 + b^2} = 2 \sum_{(a,b) \in E_R} \frac{1}{a^2 + b^2} - \sum_{(a,b) \in D_R} \frac{1}{a^2 + b^2}.$$ But $a + b$ is even if and only if one can write $(a, b) = (m - n, m + n)$, and such $m, n$ are unique. Notice also that $a^2 + b^2 = (m-n)^2 + (m+n)^2 = 2m^2 + 2n^2$, hence $a^2 + b^2 < R$ if and only if $m^2 + n^2 < R/2$. With that we get: $$\tag{6} 2 \sum_{(a,b) \in E_R} \frac{1}{a^2 + b^2} = 2 \sum_{(m,n) \in D_{R/2}} \frac{1}{(m-n)^2 + (m+n)^2} = \sum_{(m,n) \in D_{R/2}} \frac{1}{m^2 + n^2}.$$ Replacing (6) in (5), we obtain $$\sum_{(a,b) \in D_R} \frac{(-1)^{a+b}}{a^2 + b^2} = - \sum_{R/2 \le a^2 + b^2 < R} \frac{1}{a^2 + b^2},$$ where the second sum is evaluated for $a$ and $b$ integers.

Denote by $N(r)$ the number of lattice points in the open disk $x^2 + y^2 < r^2$. Along the circle with radius $r$ with $\sqrt{R/2} \le r < \sqrt{R}$, there are $N(r+0) - N(r-0)$ lattice points; each of them contribute $\frac{1}{r^2}$ in the sum (7). So we can re-write the sum as a Stieltjes integral: $$\sum_{R/2 \le a^2 + b^2 < R} \frac{1}{a^2 + b^2} = \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{1}{r^2}\,dN(r).$$ It is well-known that $N(r) = \pi r^2 + O(r)$. (Putting a unit square around each lattice point, these squares cover the disk with radius $r - 1$ and lie inside the disk with radius $r + 1$, so there their total area is between $\pi (r-1)^2$ and $\pi (r+1)^2$). By integrating by parts, $$\begin{align*} \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{1}{r^2}\,dN(r) &= \left[ N(r) \frac{1}{r^2} \right]_{\sqrt{R/2}}^{\sqrt{R}} + \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{2}{r^3} N(r)\,dr \\ &= \left[ \frac{\pi r^2 + O(r)}{r^2} \right]_{\sqrt{R/2}}^{\sqrt{R}} + 2 \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{\pi r^2 + O(r)}{r^3}\,dr \\ &= 2\pi \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{dr}{r} + O\bigl( 1/\sqrt{R} \bigr) = \pi \log 2 + O\bigl( 1/\sqrt{R} \bigr). \end{align*}$$ Therefore, $$\lim_{R \to \infty} \sum_{(a,b) \in D_R} \frac{(-1)^{a+b}}{a^2 + b^2} = - \lim_{R \to \infty} \sum_{R/2 \le a^2 + b^2 < R} \frac{1}{a^2 + b^2} = - \lim_{R \to \infty} \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{1}{r^2}\,dN(r) = -\pi \log 2.$$