When is an integral infinite help

[P3X-018]

Homework Statement

The problem is this:

Let f : R^2 -> R be

$$f(x,y) = \frac{xy}{(x^2+y^2)^2}\qquad (x,y)\neq(0,0)$$

With $f(0,0) = 0$.

Question:
Is $f$ integrable with respect to the Lebesguemeasure $m_2$ om the set $[-1,1]\times[-1,1]$?

The Attempt at a Solution

Well $f$ is integrable if and only is

$$\int_{[-1,1]^2} \vert f\vert\,dm_2(x,y) < \infty$$

The integral is infinite, and so f isn't integrable.
The thing that makes it confusing, is that in the next problem it says,
Show that both the double integrals exist

$$\int_{[-1,1]}\left(\int_{[-1,1]} f(x,y)\,dm(x)\right)dm(y)$$
$$\int_{[-1,1]}\left(\int_{[-1,1]} f(x,y)\,dm(y)\right)dm(x)$$

But how can they exist if f isn't integrable?


EDIT:
Hmm I had made an error. It did seem kinda wired that the integral of a positive would give me something like $\infty-\infty$. The integral is infinite though. The expression I had found would only hold for y > 0.

 

[mlazos]

First question : Dont you think that the integral should have two boundaries? its from [-1,1] to what? please try to be more specific so we can help

 

[P3X-018]

Actually [-1,1] is not a point it's the interval from -1 to 1. That's just another way of writting the integral of the function f(x,y) from -1 to 1 with respect to x (hence I have writtin m(x), the Lebesgue measure with respect x).

 

[HallsofIvy]

I assume that the notation "[-1,1]²" means $-1\le x\le 1$ and $-1\le y\le 1$. A function is as long as the set of points on which it is NOT continuous is a measurable set. For what points is this function not continuous?

 

[P3X-018]

I assume that the notation "[-1,1]²" means $-1\le x\le 1$ and $-1\le y\le 1$.

Yep, that's the square I'm talking about.

A function is [what?] as long as the set of points on which it is NOT continuous is a measurable set. For what points is this function not continuous?

What is the function if it's not defined on a measurable set? Integrable or measurable?
The function is measurable though, since as f is only discontinuous at the point (0,0), and we just define f(0,0)=0, and make f defined on the whole square.