Let $F:\mathbb{R}^2\to\mathbb{R}$ be
$$F(x,y)=\frac{x}{(x^2+y^2)^2}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x,y)\neq(0,0))$$
I was trying to show that $F$ is locally integrable.
[My idea]
1 ) $\ \ $ Show that F is integrable at $B(0,1)$(open ball with center at $(0,0)$ and ratio 1).
to this end, we notice that for all $\epsilon\in (0,1)$, $F$ is integrable at $B_\epsilon(0,1):=B(0,1)- B[0,\epsilon]$ with $\int_{B_\epsilon(0,1)}F(x,y)dxdy=0$ (using polar coordinates, for example)
Then $\int_{B(0,1)}F(x,y)dxdy=\lim_{\epsilon\to 0}$$\int_{B_\epsilon(0,1)}F(x,y)dxdy=0$
2 ) $ \ \ $ As $F$ is continuous in $\mathbb{R}^2-(0,0)$, then $F$ is locally integrable in $\mathbb{R}^2-(0,0)$.
For 1), and 2) , can I conclude that $F$ is locally integrable?
$F$ is locally integrable 0n $\Bbb R^2$ if $\int_K|F|<\infty$ for every compact $K\subset\Bbb R^2$. Since $$ \int_{0< x^2+y^2\le1}F(x,y)\,dxdy=\int_0^1\int_0^{2\pi}\frac{|r\cos\theta|}{r^4}\,r\,dr=4\int_0^1\frac{dr}{r^2}=\infty, $$ $F$ is not locally integrable on $\Bbb R^2$.
However, it is true that $F$ is integrable on $\{x^2+y^2\ge\epsilon\}$ for all $\epsilon>0$ and that $$ \int_{ x^2+y^2\ge\epsilon}F(x,y)\,dxdy=0. $$