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Let $f(x,y)=\dfrac{x^2-y^2}{(x^2+y^2)^2}$ prove:

$$\int_0^1\left(\int_0^1f(x,y)dy\right)dx=\frac{\pi}{4}\tag1$$

$$\int_0^1\left(\int_0^1f(x,y)dx\right)dy=\frac{-\pi}{4}\tag2$$

Why the Fubini theorem fail??

My attempt:

For $(1)$ we have:

$$\int_0^1\left(\int_0^1f(x,y)dy\right)dx=\int\int\frac{\partial}{\partial y}\left(\frac{y}{x^2+y^2}\right)=\int \frac{y}{x^2+y^2}\bigg|_{y = 0}^{y = 1}dx=\int\frac{1}{x^2+1}dx= \\ =\arctan(x)\bigg|_0^1=\frac{\pi}{4}$$

For $(2)$ we have:

$$\int_0^1\left(\int_0^1f(x,y)dx\right)dy=\int\int\frac{\partial}{\partial x}\left(\frac{-x}{x^2+y^2}\right)dxdy=-\int_0^1\frac{1}{1+y^2}dy=-\frac{\pi}{4}$$

I don't have very clear why Fubini Theorem fail. Can someone help me?

https://en.wikipedia.org/wiki/Fubini%27s_theorem (Fubini-Tonelli Theorem)

Sine of the Time
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rcoder
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    You should read the Fubini theorem carefully, looking at the assumptions. One of them will not be satisfied by your problem. – Ross Millikan Oct 06 '19 at 02:39
  • I think simply because it is not absolutely integrable? – hal4math Oct 06 '19 at 02:40
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    The function $f(x,y)$ is continuous on the unit square without the origin (0,0) but it is not bounded on the unit square because of issues near the point $(0,0)$. For example, $f(x,tx) = ((1-t^2)/(1+t^2)^2)(1/x^2)$ for $0 < x \leq 1$ and $0 < t \leq 1$. The positive function $1/x^2$ is not integrable on $(0,1]$ because of problems near $x = 0$. – KCd Oct 06 '19 at 02:52
  • Fubini's theorem does not fail. – Angina Seng Oct 06 '19 at 04:14
  • Hi! Regarding the second equality of the first equation, why the discontinuity of f(x,y) at 0 does not matter when applying what I think is some version of the fundamental theorem of calculus? – Mangostino Jun 06 '23 at 20:39

1 Answers1

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By symmetry (via Tonelli's theorem), $$\begin{align*} \int_0^1\int_0^1 |f(x,y)|\,dy\,dx &= 2\int_0^1\int_0^x \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx \\&= 2\int_0^1\frac{y}{x^2+y^2}|_{y=0}^{y=x}\,dx \\ &= \int_0^1 \frac{1}{x}\,dx \\&= \infty \end{align*}$$ Therefore, since $f(x,y)$ is not integrable over $[0,1]^2,$ Fubini's theorem doesn't apply.

Brian Moehring
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    To be clear, we can use Tonelli's theorem to evaluate the double integral as either of the integrated integrals because $(x,y)\mapsto |f(x,y)|$ is nonnegative, correct? – Math1000 Oct 06 '19 at 03:41
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    @Math1000 Yes. That isn't the use of Tonelli's theorem I referenced at the beginning (which is about changing the order of an iterated integral), but Tonelli's theorem is what allows us to use a multiple integral or iterated integrals when investigating whether a function is integrable. – Brian Moehring Oct 06 '19 at 04:20
  • Why does the first equality hold? Where does the factor of 2 come from? – Noppawee Apichonpongpan Jun 16 '21 at 18:15
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    @NoppaweeApichonpongpan Write the first integral as $$\int_0^1\int_0^1 |f(x,y)|,dy,dx = \int_0^1\int_0^x |f(x,y)|,dy,dx + \int_0^1\int_0^y |f(x,y)|,dx,dy$$ (note the use of Tonelli's theorem). Then use symmetry of $f$. – Brian Moehring Jun 19 '21 at 04:22