Proving Non-Integrability of f(x,y) on [-1,1]×[-1,1]

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Homework Statement


Prove \int\int_{[-1,1]×[-1,1]}f(x,y)dA is not Henstock Integrable.

Homework Equations


f(x,y) = \frac{xy}{(x^{2}+y^{2})^{2}}
f(x,y) = 0 if x^{2}+y^{2}=0 on the region [-1,1]×[-1,1]

The Attempt at a Solution


The only hints given is that we will not be able to solve in Maple or Mathematica by using trapezoidable interagtion techniques.

I'm assuming I would start by trying to prove that it is integrable and use contradiction to solve my answer.
 
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The Captain said:

Homework Statement


Prove \int\int_{[-1,1]×[-1,1]}f(x,y)dA is not Henstock Integrable.

Homework Equations


f(x,y) = \frac{xy}{(x^{2}+y^{2})^{2}}
f(x,y) = 0 if x^{2}+y^{2}=0 on the region [-1,1]×[-1,1]

The Attempt at a Solution


The only hints given is that we will not be able to solve in Maple or Mathematica by using trapezoidable interagtion techniques.

I'm assuming I would start by trying to prove that it is integrable and use contradiction to solve my answer.

What does it mean to be Henstock integrable? That's not a term I've ever heard.

[STRIKE]In any case, your second equation above doesn't apply. x2 + y2 ≠ 0 for any real x and y.[/STRIKE]
Edit: Wrote the above without thinking too clearly.
 
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Henstock-Kurzweil Integral is used in Real Analysis. I'm pretty sure my class is being taught material that is generally left for graduate school, however since my professor did his PhD studies on it, thinks we can handle it.

As for the second equation, we are to think of f(x,y) as a piecewise function. That way we can deal with when both x and y are zero at the same time.
 
Sorry about that incorrect comment in my previous post. I was thinking x2 + y2 ≥ 0, and that somehow that meant that x2 + y2 ≠ 0.
 
I think it's clear the problem is in 0. Try to find out what the integral is on ##([-1,1]\times[-1,1])\setminus ([-\varepsilon,\varepsilon]\times [-\varepsilon,\varepsilon])## and let ##\varepsilon\rightarrow 0##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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