6
$\begingroup$

I am trying to come up with a measurable function on $[0,1]^2$ which is not integrable, but such that the iterated integrals are defined and unequal.

Any help would be appreciated.

2 Answers 2

6

Consider the double integrals $I:=\int_0^1\int_0^1{y-x\over(2-x-y)^3}\ dy\ dx\ ,\qquad J:=\int_0^1\int_0^1{y-x\over(2-x-y)^3}\ dx\ dy\ .$ Then $\int_0^1{y-x\over(2-x-y)^3}\ dy={y-1\over(2-x-y)^2}\Biggr|_{y=0}^1={1\over(2-x)^2}\ .$ It follows that $I=\int_0^1 {dx\over(2-x)^2}={1\over 2-x}\Biggr|_0^1={1\over2}\ .$ Similarly you get $J=-{1\over2}\ne I$.

  • 0
    How do you show that \int_0^1\int_0^1 | {y-x\over(2-x-y)^3 }\ | dx\ dy\ < \infty ? Is there any obvious way that doesn't require much calculations?2018-05-25
2

$ \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dy\,dx \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dx\,dy $

Obviously either of these is $-1$ times the other and if this function were absolutely integrable, then they would be equal, so their value would be $0$. But one is $\pi/2$ and the other is $-\pi/2$, as may be checked by freshman calculus methods.