Can We always change the order of integration in double integrals ?
Calculus Question - change order of integration
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1You can always change the order of integration, but that does not mean values will coincide! If the hypothesis of Fubini's theorem is satisfied the values will coincide. – 2012-11-25
1 Answers
Here is a counterexample that shows the value might not coincide.
Consider the function $ \frac{x^2-y^2}{(x^2+y^2)^2} $ A $y$-primitive for this on $[1,\infty)$ is $ \frac{y}{x^2+y^2} $ which simplifies to $-\frac{1}{1+x^2},$ when evaluated from $1$ to $\infty$.
Knowing this, and the fact that $\int \frac{1}{1+x^2}=\arctan(x)+C,$ we get $ \int_1^\infty \int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2} dydx=\int_1^\infty -\frac{1}{1+x^2}=-\frac{\pi}{4}. $
Now changing the order, we get $ \int_1^\infty \int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2} dxdy=-\int_1^\infty \int_1^\infty \frac{-x^2+y^2}{(x^2+y^2)^2} dxdy=\int_1^\infty \frac{1}{1+y^2}=\frac{\pi}{4}. $
There are theorem that can garantee the change will provide the same answer, namely Fubini's. As Glebovg pointed out, you can find this example and some other theorems here.
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1@glebovg Classics are what they are sometimes. That's the first example I remember seeing of a situation where changing the order of integration leads to a different answers. At the time, I was a bit awed, and wonder how one could come up with an example like that. Then my teacher told me that the goal was to keep it simple, try to get something as "insignifiant" as a change of sign. Now what do you do to get that, you look for a fonction with some kind of symmetry that has f(x,y)=-f(y,x). Then you want to make sure the first integral gives the same result. All of this to say, I remembered – 2012-11-25