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I have a function defined as follows: $f(x,y)= \dfrac{x^2-y^2}{\left(x^2+y^2\right)^2}$, if $(x,y)\neq (0, 0)$ and $f(x,y)=0$ if $(x,y)=(0,0)$. Now, $$\int_0^1\int_0^1 f(x,y)\,\text{d}x~\text{d}y=-\frac{\pi}{4}$$ and $$\int_0^1\int_0^1 f(x,y)\,\text{d}y~\text{d}x=\frac{\pi}{4}.$$ The question I have is this: why does this not contradict Fubini's theorem?

thanks.

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    Read a statement of Fubini's theorem. Go through its hypotheses. Work out which is not satisfied.2011-11-24
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    For crying out loud... the analysis of this **exact** function is on the wikipedia page... http://en.wikipedia.org/wiki/Fubini%27s_theorem#Rearranging_a_conditionally_convergent_iterated_integral2011-11-24
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    @TheChaz: For crying out loud...did you read my question? I asked why Fubini's theorem is NOT refuted, and not why it doesn't hold?2011-11-24
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    @Chris What do you mean by "Fubini's theorem is NOT refuted"? The theorem is not refuted because it is not even applicable for this example. And that is because the premises or hypotheses of the theorem are not met.2011-11-24
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    @Srivatsan: Well that was the question asked. Moreover, at the linked page given by The Chaz, after, giving the example, they said that...Fubini's theorem, says that...so if the conditions are not satisfied, why are they applying the it?2011-11-24
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    Why the downvotes? What is wrong with the question?2011-11-24
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    I didn't downvote. [Fubini's theorem](http://en.wikipedia.org/wiki/Fubini's_theorem) says: *if* $\int_{A \times B} |f(x,y)|\,d(x,y)\lt\infty$ *then* the double integral $\int_{A \times B} f(x,y)\,d(x,y)$ is equal to the iterated integrals you write. Here you have two iterated integrals that aren't equal, so you can conclude from Fubini's theorem that the double integral isn't finite. And indeed, as Michael's answer shows, the first double integral I mentioned is infinite. No contradiction here, but a cautionary example showing that *some* hypotheses are necessary to get a Fubini-like theorem.2011-11-24
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    To supplement t.b.'s answer, let's note that the theorem is a conditional statement "if P, then Q". P is about finiteness; Q is about integrals being equal. The contra positive states (again, roughly) that if the integrals aren't equal, then there was an infinity to start out with! Also, I haven't voted on this question.2011-11-24
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    @All: Thanks very much for taking the time to make me sort out my confusion.2011-11-24

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Because the integrals of the positive and negative parts are both infinite.

If $x>y$ then $(x^2-y^2)/(x^2+y^2)^2$ is positive; if $x

It's the same thing with infinite series: If the sum of the positive terms is $+\infty$ and the sum of the negative terms is $-\infty$, and the series adds up to some finite number, then you can make it add up to a different finite number by rearranging the terms.

If a function is always positive, and its integral is $\infty$, the rearranging it won't change the fact that its integral is $\infty$. Therefore, if one looks at the absolute value $$ \int_0^1\int_0^1 \left|\frac{x^2-y^2}{(x^2+y^2)^2}\right| \;dy\;dx, $$ it will remain $+\infty$ no matter how you rearrange it. And that is precisely the circumstance in which Fubini's theorem does not apply.

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    Thanks for your answer. Unless they mean the same, my question was why Fubin's theorem is not refuted? and not why Fubini does not apply?2011-11-24
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    I say that if the president is me, then I will kill all the ET, but I am not, so the *theorem* does not apply, and hence that there are some ET does not refute this *theorem*. Regards here. P.S. I downvoted not.2011-11-24
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    @Chris: Dear Chris, What distinction are you trying to draw between "Fubini's theorem is not refuted" and "Fubini's theorem does not apply"? As you note in your question, the conclusion of Fubini's theorem does not hold in this case. As Michael notes, the hypotheses of Fubini's theorem don't hold either (which he expresses via the phrase "Fubini's theorem does not apply"). This exactly answers your question: Fubini's theorem is not refuted because in this example because its hypotheses do not hold. What more do you want? Regards,2011-11-24
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    @Chris : Fubini's theorem is not refuted because it is not applicable in this situation.2011-11-24
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    @Chris: In order for a theorem to be refuted, you must have an example in which (i) the hypotheses of the "theorem" are true; and (ii) the conclusion of the "theorem" is false. Having an example where the hypotheses and conclusion are both false, or an example where the hypotheses are false and the conclusion is true, does not constitute a refutation.2011-11-24
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    @Arturo: Thanks very much for your comment. I'm alright now.2011-11-24