integral over a subset of $\mathbb {R}^2$ is not defined while...

Question

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?

Answer 1

$f(x,y)$ is positive in the first and third quadrants and negative in the second and fourth. One would need to show that the integral over the first and third quadrants $\text{is }{+\infty}$, and then by an easy symmetry argument, it follows that the integral over the second and fourth quadrants $\text{is }{-\infty}$. Let's look at $$ \iint_R f(x,y)\,d(x,y) $$ where $R$ is a small quarter-disk of radius $r_1\le 1$ in the first quadrant with the center of the whole disk at $(0,0)$. $$ \iint_R f(x,y)\,d(x,y) = \int_0^{\pi/2} \int_0^{r_1} \frac{r^2\cos\theta\sin\theta}{r^4} r\,dr\,d\theta = \int_0^{\pi/2} \sin(2\theta)\,d\theta\cdot\int_0^{r_1} \frac{dr} r = \infty. $$

Answer 2

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 $s\to0$.

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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 $s\to0$.