Different results after changing the order of integration with constant limits (Failure of Fubini's theorem)

Question

I have the following question

$I_{1}=\int _{0}^{1}\int _{0}^{1}\ \frac{(x-y)}{(x+y)^{3}}\ dy\,dx$

Evaulating the above I get $I_{1}=0.5$

Now if I switch the order of integration

$I_{2}=\int _{0}^{1}\int _{0}^{1}\ \frac{(x-y)}{(x+y)^{3}}\ dx\,dy$

I get $I_{2}=-0.5$

Why is the value negative after changing the order? Shouldn't the result be same for constant limits (as the region of space is the same for both the integrals) ?

Answer 1

As pointed out by @Jack D'Aurizio, the answer lies in the below quote from Wikipedia

If the above integral of the absolute value is not finite, then the two iterated integrals may have different values.

Failure of Fubini's theorem for non-integrable functions

Fubini's theorem tells us that (for measurable functions on a product of σ-finite measure spaces) if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect to x and then with respect to y, we get the same result as if we integrate first with respect to y and then with respect to x. The assumption that the integral of the absolute value is finite is "Lebesgue integrability", and without it the two repeated integrals can have different values.

A simple example to show that the repeated integrals can be different in general is to take the two measure spaces to be the positive integers, and to take the function f(x,y) to be 1 if x=y, −1 if x=y+1, and 0 otherwise. Then the two repeated integrals have different values 0 and 1.

Another example is as follows for the function:

${\displaystyle {\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}=-{\frac{\partial ^{2}}{\partial x\partial y}}\arctan(y/x).}$

The iterated integrals

${\displaystyle \int _{x=0}^{1}\left(\int _{y=0}^{1}{\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,{\text{d}}y\right)\,{\text{d}}x={\frac{\pi }{4}}}$

and ${\displaystyle \int _{y=0}^{1}\left(\int _{x=0}^{1}{\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,{\text{d}}x\right)\,{\text{d}}y=-{\frac{\pi }{4}}}$

have different values. The corresponding double integral does not converge absolutely (in other words the integral of the absolute value is not finite):

${\displaystyle \int _{0}^{1}\int _{0}^{1}\left|{\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\right|\,{\text{d}}y\,{\text{d}}x=\infty.}$