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I am going over my lecture's notes in preparation for exam and I saw something a bit strange I would like someone to explain how it is possible.

Look at the function $f(x,y) = x^4-6x^2y^2+y^4$

if we convert it to polar coordinates, we will get $f(r,\theta)=r^4\cos(4\theta)$

the weird thing about this function, is that it has 2 different integrals over the same area.

What I mean: if we calculate the integral $$ \lim_{n \to \infty}\int_{0}^{2\pi} \int_{0}^{n} r^5\cos(4\theta)dr d\theta$$ we will see it is equal to zero.

However, if we calculate $$\lim_{n \to \infty}\int_{-n}^{n} \int_{-n}^{n} x^4-6x^2y^2+y^4dxdy = \lim_{n \to \infty}\int_{-n}^{n} \frac{2n^5}{5}-4n^3y^2+2ny^4 dy=\lim_{n \to \infty} -\frac{16}{15}n^6 = -\infty$$

How come the integral is so different based just on how we choose to represent the same area? (it is $\mathbb R^2$ in both cases. first it is a giant circle, second case it is a giant square)

So if asked what is $$\iint_{\mathbb R^2} x^4-6x^2y^2+y^4$$ is the correct thing to say that it does not exist?

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    $\begingroup$ $f$ is not integrable. The integral $\iint_{\mathbb{R}^2} f(x,y)\,dx\,dy$ does not exist. You exhaust the plane by disks (centered at $0$) in the one case, and by squares in the other. The difference between the square $[-n,n]^2$ and the disk $x^2 + y^2 \leqslant n^2$ consists of four "curved triangles" on which the integrand is predominantly negative. Exhaust the plane with tilted squares, $\lvert x\rvert + \lvert y\rvert \leqslant n$, then the limit might be $+\infty$. $\endgroup$ – Daniel Fischer Jun 18 '14 at 19:04
  • $\begingroup$ @DanielFischer $f$ should be integrable over a finite subset of $\mathbb R^2$ $\endgroup$ – kleineg Jun 18 '14 at 19:09
  • $\begingroup$ But the geometric interpretation is appreciated. $\endgroup$ – kleineg Jun 18 '14 at 19:10
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    $\begingroup$ @kleineg Sure, it's continuous, hence locally integrable. But it's not integrable over the entire plane. $\endgroup$ – Daniel Fischer Jun 18 '14 at 19:13
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The integral is unbounded when integrated over the entire plane of $\mathbb R^2$, what you are doing by changing the variables is analogous to changing the order of the terms in an non-converging summation, you can get more than one "answer" but in fact none of them are valid.

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    $\begingroup$ Is this because some subsequences diverge to $\infty$ and some to $-\infty$ so it violates Fubini/Tonelli? $\endgroup$ – Avraham Jun 18 '14 at 19:16
  • $\begingroup$ That is precisely it. $\endgroup$ – kleineg Jun 18 '14 at 19:22
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This integral does not converge. To illustrate, we can see that $$\lim_{n \to \infty} \int_{-n}^{n}x \, \text dx = \lim_{n \to \infty} \left( \frac{n^2}{2} - \frac{(-n)^2}{2} \right) = \lim_{n \to \infty} 0 = 0$$ even though $\int_{-\infty}^{\infty} x \, \text dx$ does not exist. What was just computed is called the Cauchy principal value, which assigns values to integrals which would otherwise be undefined.

Your case is similar, except that in 2D the limit of the expanding domain can be taken in many ways, which give different answers. Furthermore, the fact that the integral taken in two equivalent ways give different results is sufficient to state that it is undefined.

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