Let's denote:
$$Q_k := (k,k+1)\times(k,k+1)$$ $$V_k :=(k,k+1)\times(-2k,-k)$$
let be $$F(x,y) = \sum_{k=1}^{+\infty} \frac{1}{x^2+y^2}(\mathbf{1}_{Q_k}(x,y)-\mathbf{1}_{V_k}(x,y))$$
Verify that $F$ is measurable and limited in $\mathbb{R^2}$ showing a lower and upper bound.
Denoted $H_+ := \{(x,y)\in \mathbb{R}^2 : y>0\} $ $H_- := \{(x,y)\in \mathbb{R}^2 : y<0\} $, establish whether $F\in L^1(H_+)$ and whether $ F\in L^1(H_-)$
Calculate, if it exist, $\int_{R^2} F(x,y)dxdy$
I'll post an answer given by a friend of mine on the main problem: the integration on $H_+$ or $H_-$
Let's consider $H_+$
$$\int_{H_+} F(x,y)dm = \sum_{k=1}^{\infty} \int_{Q_k} F(x,y)dm \le \sum_{k=1}^{\infty}\int_{k}^{k+1}\int_{k}^{k+1}\frac{1}{x^2+y^2}dxdy$$
Now, because $x \in (k,k+1)$,
$$ k < x \ \ \Rightarrow \ \ k^2 < x^2 \ \ \Rightarrow \ \ \ 2k^2<x^2 + y^2 \ \ \Rightarrow \ \ \frac{1}{x^2 + y^2} < \frac{1}{2k^2}$$
So we can conclude
$$\sum_{k=1}^{\infty}\int_{k}^{k+1}\int_{k}^{k+1}\frac{1}{x^2+y^2}dxdy < \sum_{k=1}^{\infty} \frac{1}{2k^2} < \infty$$