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Consider the function

$$ f(x,y)= \frac{x^2-y^2}{(x^2+y^2)^2}. $$

Now, if you evaluate the integral

$$ \int_{0}^{1}\int_{0}^{1}f(x,y)dydx = \frac{\pi}{4},$$

and if you consider the other order, you get

$$ \int_{0}^{1}\int_{0}^{1}f(x,y)dxdy = -\frac{\pi}{4}. $$

So, the iterated integrals exist, but the double integral does not.

My question: Why is $f$ not integrable? It continuous except measure zero set.

PSUN
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    It is not bounded, and both the positive part and negative part have a divergent integral. Hence, this $f$ Riemann integrable (in the generalized sense), but not Lebesgue integrable. Note that the theorem you are evoking doesn't apply since $f$ is not bounded. – Crostul Oct 03 '16 at 09:19
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    @Crostul: Why not put that as an answer? – Hans Lundmark Oct 03 '16 at 10:27

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