4

I have to prove that the following integral does not exist:

$$\iint \limits _{[0,1] \times [0,1]} \frac{x^2-y^2}{(x^2 + y^2)^2}\,\mathrm dx\,\mathrm dy .$$

I think I can use Fubini's Theorem, ie. if I show that iterated integrals are not equal to each other, then the corresponding double integral can't be integrable, because otherwise it would be "counter example" to Fubini's Theorem. Is that a good idea?

Alex M.
  • 35,207
luka5z
  • 6,359
  • 1
    Some remarks: (1) Your integrand changes sign if you swap $x$ and $y$, so by symmetry, the integral must be zero if it exists. (2) You can restrict your attention to the set ${(x,y);x\geq0,y\geq0,x^2+y^2\leq1}$ (the integrand is nice outside it) and use polar coordinates. (3) It seems that the iterated integrals don't exist (I haven't checked) but they should if the integral did exist. (4) The integral exists in a principal value sense and is zero then, but this is probably not what you want. – Joonas Ilmavirta Jul 04 '15 at 09:31
  • I believe $\int_{x=0}^1\left(\int_{y=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}dy\right)dx=\pi/4$ and $\int_{y=0}^1\left(\int_{x=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}dx\right)dy=-\pi/4$. So from Fubini theorem double integral doesnt exist. Is that right? Note that $\frac{x^2-y^2}{(x^2+y^2)^2} = -\frac{\partial^2}{\partial x\partial y}\arctan(y/x)$ – luka5z Jul 04 '15 at 10:08
  • 1
    I can't do those integrals in my head, but you can always ask if Wolfram alpha agrees with you. http://www.wolframalpha.com/ – Joonas Ilmavirta Jul 04 '15 at 10:13
  • I didnt want you to check integrals, but the general idea. If I proved that those integrals exist and are not equal, then the double integral doesnt exist (from Fubini). Is that caorrect reasoning? – luka5z Jul 04 '15 at 11:00
  • 1
    Yes, the reasoning is perfectly fine. – Joonas Ilmavirta Jul 04 '15 at 11:01

2 Answers2

4

The region $\{(x,y): x\ge y\ge 0,\;x^2+y^2\le 1\}$ is contaied in $[0,1]\times[0,1]$, and the integrand is nonnegative there. Furthermore, we can easily parametrize it in polar coordinates:

$$\int_0^{\pi/4}\int_0^1r\frac{r^2\cos2\theta}{r^4}drd\theta$$

But $1/r$ is not integrable at $r\in(0,1]$.

ajotatxe
  • 65,084
2

What you wrote is not exactly right. The function $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ is not integrable with respect to the Lebesgue measure on the unit square $(0,1)\times(0,1)$ (so it does not fulfill the hypothesis of Fubini's theorem), but the integral $$ I=\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx$$ does exist, since: $$ \int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dy = \frac{1}{1+x^2} $$ hence: $$ I = \int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}.$$ However, it is straightforward to check that: $$ \frac{\pi}{4}=\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx\neq \int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy = -\frac{\pi}{4}. $$ The integral that is not defined is: $$ \iint_{(0,1)^2} f(x,y) \,d\mu $$ where $\mu$ is the Lebesgue measure on the unit square.

Jack D'Aurizio
  • 353,855