integral over a subset of $\mathbb {R}^2$ is not defined while... consider the function $f(x,y)=\frac{xy}{(x^2+y^2)^2}$, we can see by some easy calculation that $\int_{-1}^1\int_{-1}^1 f(x,y)\,dx\, dy$ and $\int_{-1}^1\int_{-1}^1 f(x,y)\,dy\, dx$ exist and equals $0$.
but the function is not integrable over the square $-1<x<1 , -1<y<1 $.i must prove this.i think it is because in a small neighborhood of $0$,the function grows really fast.can some one help me to write this imagination in detail?

Note that $f(x,y)$ is homogeneous of order $-2$. That is $f(ax,ay)=a^{-2}f(x,y)$. If $f$ is not identically $0$, then the integral of $|f|$ over a unit circle is $I\ne0$. In fact,
$$
\begin{align}
I
&=\int_0^{2\pi}|\cos(\theta)\sin(\theta)|\,\mathrm{d}\theta\\
&=4\int_0^{\pi/2}\frac12\sin(2\theta)\,\mathrm{d}\theta\\
&=\left[-\cos(2\theta)\vphantom{\int}\right]_0^{\pi/2}\\[6pt]
&=2
\end{align}
$$
Therefore,
$$
\begin{align}
\int_{s\le|(x,y)|\le1}|f(x,y)|\,\mathrm{d}x\,\mathrm{d}y
&=\int_s^1\frac2{r^2}\,r\,\mathrm{d}r\\
&=2\log\left(\frac1s\right)\\[6pt]
&\to\infty
\end{align}
$$
as $s\to0$.

I had deleted this because Michael Hardy had answered earlier, and I didn't think that mentioning the homogeneity was enough to add for another answer. However, I realized that the idea that the integral of $f$ exists is the two-dimensional analog of the Cauchy Principal Value. That is, the integral of $f$ around a unit circle is
$$
\begin{align}
\int_0^{2\pi}\cos(\theta)\sin(\theta)\,\mathrm{d}\theta
&=\int_0^{2\pi}\frac12\sin(2\theta)\,\mathrm{d}\theta\\
&=\left[-\frac14\cos(2\theta)\right]_0^{2\pi}\\[6pt]
&=0
\end{align}
$$
Therefore,
$$
\begin{align}
\int_{s\le|(x,y)|\le1}f(x,y)\,\mathrm{d}x\,\mathrm{d}y
&=\int_s^1\frac0{r^2}\,r\,\mathrm{d}r\\
&=0\log\left(\frac1s\right)\\[6pt]
&\to0
\end{align}
$$
as $s\to0$.
