# Not integrable although iterated integrals are equal

How can I show that the function

$$f=\begin{cases} 0 & (x,y)=(0,0)\\\frac{xy}{(x^2+y^2)^2} & \mbox{else}\end{cases}$$ is not Lebesgue-integrable, although the iterated integrals exist and are equal: $$\int_{-1}^{1}\int_{-1}^{1}f(x,y)dydx=\int_{-1}^{1}\int_{-1}^{1}f(x,y)dxdy?$$

• Normally, the way this happens is that, although the positive and negative parts of the function manage to cancel out in the iterated integrals, they are both infinite overall. So you want to compute the Lebesgue integral $\int_{[0,1]\times [0,1]} f^{+}\,dA$ and similarly the one for $f^{-}$. Recall that the definition of the Lebesgue integral is that $\int f = \int f^+ - \int f^-$, provided that at most one of the two subintegrals is infinite. The integrand has a clear rotational symmetry, which leads me to suspect that $f^+$ and $f^-$ will both be infinite if either is. – Carl Mummert Jun 4 '16 at 13:44

## 1 Answer

Hint. As $(x,y) \to (0,0)$, using polar coordinates we have, $$f(x,y) \sim \frac{\sin \theta\cos \theta}{r^2}$$ which is not integrable as $r \to 0$.

• Ok thank you, I will have a look on how to use polar coordinates.. – Tesla Jun 4 '16 at 13:40
• You are welcome. – Olivier Oloa Jun 4 '16 at 13:44
• By the change of variable $x=r\cos \theta,\,y=r\sin \theta$, the differential element $dx\,dy$ becomes $r\:dr\,d\theta$, the initial set of integration $[-1,1]\times [-1,1]$ becomes $[0,\pi]\times [0,1]$ and the function $f(x,y)$ becomes $\frac{\sin \theta\cos \theta}{r^2}$. Thus the initial integral becomes $\int_{0}^{\pi}\sin\theta\cos\theta \, d\theta\int_{0}^1\frac{1}{r}\, dr$. – Olivier Oloa Jun 5 '16 at 17:42
• Thanks for the detailed explanation, got it now... – Tesla Jun 5 '16 at 17:45
• @Sigma It is $[0,2\pi]\times [0,1]$, sorry for what I wrote in my comment above. But this does not change the conclusion $\int_0^1 \frac1r \:dr$ is still not convergent. Thanks. – Olivier Oloa Jun 5 '16 at 19:07