# very strange phenomenon $f(x,y)=x^4-6x^2y^2+y^4$ integral goes wild

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?

• $f$ 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 $+\infty$. – Daniel Fischer Jun 18 '14 at 19:04
• @DanielFischer $f$ should be integrable over a finite subset of $\mathbb R^2$ – kleineg Jun 18 '14 at 19:09
• But the geometric interpretation is appreciated. – kleineg Jun 18 '14 at 19:10
• @kleineg Sure, it's continuous, hence locally integrable. But it's not integrable over the entire plane. – Daniel Fischer Jun 18 '14 at 19:13

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.
• Is this because some subsequences diverge to $\infty$ and some to $-\infty$ so it violates Fubini/Tonelli? – Avraham Jun 18 '14 at 19:16
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.