Define $\mathcal{E}_{R}=\left\{(a, b) \in \mathbb{Z}^{2} \backslash\{(0,0)\}: a^{2}+b^{2}<R\right.$ and $a+b$ is even $\}$. Then $\sum_{(a, b) \in \mathcal{D}_{R}} \frac{(-1)^{a+b}}{a^{2}+b^{2}}=2 \sum_{(a, b) \in \mathcal{E}_{R}} \frac{1}{a^{2}+b^{2}}-\sum_{(a, b) \in \mathcal{D}_{R}} \frac{1}{a^{2}+b^{2}}$. 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 $a^{2}+b^{2}=(m-n)^{2}+(m+n)^{2}=2 m^{2}+2 n^{2}$, hence $a^{2}+b^{2}<R$ if and only if $m^{2}+n^{2}<R / 2$. With that we get: $2 \sum_{(a, b) \in \mathcal{E}_{R}} \frac{1}{a^{2}+b^{2}}=2 \sum_{(m, n) \in D_{R / 2}} \frac{1}{(m-n)^{2}+(m+n)^{2}}=\sum_{(m, n) \in D_{R / 2}} \frac{1}{m^{2}+n^{2}}$. Replacing this in the earlier expression, we obtain $\sum_{(a, b) \in \mathcal{D}_{R}} \frac{(-1)^{a+b}}{a^{2}+b^{2}}=-\sum_{R / 2 \leq a^{2}+b^{2}<R} \frac{1}{a^{2}+b^{2}}$ 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 $\sqrt{R / 2} \leq r<\sqrt{R}$, there are $N(r+0)-N(r-0)$ lattice points; each of them contribute $\frac{1}{r^{2}}$ in the sum. So we can re-write the sum as a Stieltjes integral: $\sum_{R / 2 \leq a^{2}+b^{2}<R} \frac{1}{a^{2}+b^{2}}=\int_{\sqrt{R / 2}}^{\sqrt{R}} \frac{1}{r^{2}} \mathrm{~d} N(r)$. It is well-known that $N(r)=\pi r^{2}+O(r)$. By integrating by parts, $\int_{\sqrt{R / 2}}^{\sqrt{R}} \frac{1}{r^{2}} \mathrm{~d} N(r) =\left[\frac{1}{r^{2}} N(r)\right]_{\sqrt{R / 2}}^{\sqrt{R}}+\int_{\sqrt{R / 2}}^{\sqrt{R}} \frac{2}{r^{3}} N(r) \mathrm{d} r =\left[\frac{\pi r^{2}+O(r)}{r^{2}}\right]_{\sqrt{R / 2}}^{\sqrt{R}}+2 \int_{\sqrt{R / 2}}^{\sqrt{R}} \frac{\pi r^{2}+O(r)}{r^{3}} \mathrm{~d} r =2 \pi \int_{\sqrt{R / 2}}^{\sqrt{R}} \frac{\mathrm{d} r}{r}+O(1 / \sqrt{R})=\pi \log 2+O(1 / \sqrt{R})$. Therefore, $\lim _{R \rightarrow \infty} \sum_{(a, b) \in \mathcal{D}_{R}} \frac{(-1)^{a+b}}{a^{2}+b^{2}}=-\lim _{R \rightarrow \infty} \sum_{R / 2 \leq a^{2}+b^{2}<R} \frac{1}{a^{2}+b^{2}}=-\lim _{R \rightarrow \infty} \int_{\sqrt{R / 2}}^{\sqrt{R}} \frac{1}{r^{2}} \mathrm{~d} N(r)=-\pi \log 2