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: $$\int_{-1}^{1}\int_{-1}^{1}f(x,y)dydx=\int_{-1}^{1}\int_{-1}^{1}f(x,y)dxdy?$$

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    $\begingroup$ 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. $\endgroup$ Jun 4, 2016 at 13:44
  • $\begingroup$ How does one show that the iterated integrals are equal? $\endgroup$
    – user135520
    Mar 30, 2020 at 15:07

1 Answer 1


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 $r \to 0$.

  • $\begingroup$ Ok thank you, I will have a look on how to use polar coordinates.. $\endgroup$
    – Tesla
    Jun 4, 2016 at 13:40
  • $\begingroup$ You are welcome. $\endgroup$ Jun 4, 2016 at 13:44
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    $\begingroup$ 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 $\int_{0}^{\pi}\sin\theta\cos\theta \, d\theta\int_{0}^1\frac{1}{r}\, dr$. $\endgroup$ Jun 5, 2016 at 17:42
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    $\begingroup$ Thanks for the detailed explanation, got it now... $\endgroup$
    – Tesla
    Jun 5, 2016 at 17:45
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    $\begingroup$ @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. $\endgroup$ Jun 5, 2016 at 19:07

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