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?


  • 5
    $\begingroup$ Read a statement of Fubini's theorem. Go through its hypotheses. Work out which is not satisfied. $\endgroup$ Nov 24, 2011 at 0:49
  • 16
    $\begingroup$ For crying out loud... the analysis of this exact function is on the wikipedia page... en.wikipedia.org/wiki/… $\endgroup$ Nov 24, 2011 at 1:23
  • 3
    $\begingroup$ @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. $\endgroup$
    – Srivatsan
    Nov 24, 2011 at 2:37
  • 6
    $\begingroup$ I didn't downvote. 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. $\endgroup$
    – t.b.
    Nov 24, 2011 at 3:00
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 24, 2011 at 3:47

1 Answer 1


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 $x<y$, it is negative. The boundary is the diagonal line $x=y$, dividing the square into two halves. Integrating this function over the half where the function is positive yields $+\infty$; integrating over the other half yields $-\infty$.

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.

  • $\begingroup$ 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? $\endgroup$
    – Chris
    Nov 24, 2011 at 2:17
  • 7
    $\begingroup$ @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, $\endgroup$
    – Matt E
    Nov 24, 2011 at 3:27
  • 2
    $\begingroup$ @Chris : Fubini's theorem is not refuted because it is not applicable in this situation. $\endgroup$ Nov 24, 2011 at 4:27
  • 5
    $\begingroup$ @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. $\endgroup$ Nov 24, 2011 at 5:15
  • 3
    $\begingroup$ @Arturo: Thanks very much for your comment. I'm alright now. $\endgroup$
    – Chris
    Nov 24, 2011 at 7:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .