Show that the function $f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}$ is not Lebesgue integrable on the unit square $[0,1]^2$.

Question

Show that the function $$f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}$$ is not Lebesgue integrable on the unit square $[0,1]^2$.

It's been asked here before (Is the function $f(x,y) = \frac{x^2 - y^2}{(x^2+y^2)^2}$ lebesgue integrable in $[0,1]\times[0,1]$?). However I don't really understand the answers.

I'm not very comfortable with product measures, but it seems Fubini's theorem might be useful.

Edit: I was thinking about something along the same lines as in this example from Wikipedia. Is that correct?

Answer 1

A different thing you can try for this particular integral is the following. First, note that since $x^2=(-x)^2$, we have $x^2-y^2$ is invariant under flipping across the $x$ or $y$ axes. Thus its Lebesgue integrable iff the integral over $[-1,1]^2$ is finite. And for this it suffices to show that the (unsigned) integral over a small ball around 0, $B(0,r)$ is infinite.

For integrals on balls around singularities, you can count the degree of the polynomial as a heuristic. On top, you have a degree 2 polynomial, and at the bottom you have a degree 4 polynomial. Therefore, at best, you have a pole of order 2, and in $\mathbb R^d$, only poles of order $<d$ are integrable. This is something you can verify for the simpler integral $\int_{B(0,r)} \frac{d\vec x}{|\vec x|^{d}}$ using radial coordinates, and you can prove each case via calculations similar to the other answer.

---

Regarding your edit, yes, you can absolutely do it that way.

Fubini's Theorem says that if a function $f(x,y)$ is Lebesgue integrable on the product space, then the two iterated integrals must be the same. Wikipedia uses the contrapositive- they show that the two iterated integrals are different, which implies that $f(x,y)= -\partial_x\partial_y \arctan(y/x)$ is not Lebesgue integrable.

Well, they didn't exactly show it. But its not that hard, if you knew that $f$ miraculously had this formula! My experience is that this is quite rare, and the other methods work more generally.

But lets work through it anyway. You can proceed by first computing derivatives and working backwards: since $$\int_0^1 -\partial_x\partial_y \arctan(y/x) dx = -\partial_y \arctan(y/x)|_{x=0}^1 = \left.\frac{-1}x\frac1{y^2/x^2 + 1}\right|_{x=0}^1 = \left.\frac{-x}{y^2+x^2}\right|_{x=0}^1 = \frac{-1}{y^2+1}$$ we have $$\int_0^1 \int_0^1 -\partial_x\partial_y \arctan(y/x) dxdy =\int_0^1\int_0^1 \frac{x^2 - y^2}{(x^2+y^2)^2} dxdy = \int_0^1 \frac{-1}{y^2 + 1} dy < 0$$ On the other hand - $$\int_0^1 -\partial_x\partial_y \arctan(y/x) dy = -\partial_x \arctan(y/x)|_{y=0}^1 = -1\times \left.\frac{-y}{x^2}\frac1{y^2/x^2 + 1}\right|_{y=0}^1 = \left.\frac{y}{y^2+x^2}\right|_{y=0}^1 = \frac{1}{x^2+1}$$ which implies $$\int_0^1 \int_0^1 -\partial_x\partial_y \arctan(y/x) dydx = \int_0^1\int_0^1 \frac{x^2 - y^2}{(x^2+y^2)^2} dydx = \int_0^1 \frac{+1}{x^2 + 1} dy > 0$$ A related observation- note that $f(x,y) = -f(y,x)$. Therefore

$$ \int_0^1 \int_0^1 f(x,y) dxdy = \int_0^1 \int_0^1 -f(y,x) dxdy$$ But in the right hand side, $y$ and $x$ are dummy variables that are integrated over the same region. So we can switch all $y$s with all $x$s to get $$ \int_0^1 \int_0^1 f(x,y) dxdy = - \int_0^1 \int_0^1 f(x,y) dy dx$$ This means that if $f$ is integrable, its integral must be $0$. So for example, $\sin\frac{x^2 - y^2}{(x^2+y^2)^2}$ is Lebesgue integrable on $[0,1]^2$ with integral 0.

Answer 2

Although we already have a couple of answers to this question, I am just making some comments that might be helpful. First of all, we recall the definition of Lebesgue integrable. The definition says a measurable function $f$ is integrable on a set(measurable) $A$ if $\int\limits_{A}|f|d\mu<\infty$.

Note that in general, it's difficult to work with product measures. Basically, we prove this exercise by contradiction. By the way, the given function is continuous, and hence measurable. Therefore, the integration with respect to product measure makes sense.

Now, Suppose the function is integrable on $[0,1]^2$. Hence, Fubini's theorem must hold. That means the integration with respect to product measure and the two iterated integrals with different ordering must exist and they are equal to some finite number.

Unfortunately, the iterated integrals are $\infty$ in this case. To prove this claim, we just change our coordinate systems. This part is easy. To make sure that the two iterated integrals are $\infty$, you can simply use the Monotone convergence theorem. At some point you need to prove that $\int_{(0,1]} 1/x = \infty$. Try to prove this thing at first.

Since we could not show the assertions of Fubini's Theorem, the function is not integrable. I hope this helps. I did not show any computations, because you already have some the answers with explicit computations.

Answer 3

Consider the sector $A_\delta = \{(x,y): \delta \leqslant \sqrt{x^2+y^2}\leqslant 1,\, 0 \leqslant \arctan(y/x) \leqslant \pi/2\} \subset [0,1]^2$ and the indicator function $\chi_{A_\delta}(x,y) = \begin{cases}1, & (x,y) \in A_\delta, \\ 0, & (x,y) \notin A_\delta \end{cases}$

For all $(x,y) \in [0,1]^2$ we have

$$|f(x,y)| = \left|\frac{x^2-y^2}{(x^2+y^2)^2} \right| \geqslant |f(x,y)|\chi_{A_\delta}(x,y) ,$$

and, changing $(x,y)$ to polar coordinates $(r,\theta)$ we have

$$\int_{[0,1]^2} |f| \geqslant \int_{[0,1]^2} |f| \chi_{A_\delta} = \int_{0}^{\pi/2}\int_\delta^1 \frac{r^2|\cos^2 \theta - \sin^2\theta |}{r^4}r\, dr \, d\theta \\ $$

The integral on the RHS is over a region where the integrand is continuous and may be evaluated as an iterated Riemann integral.

Thus,

$$\int_{[0,1]^2} |f| \geqslant \int_\delta^1 \frac{dr}{r} \int_0^{\pi/2}|\cos^2 \theta - \sin^2 \theta| \, d\theta = \ -\log\delta\int_0^{\pi/2} |\cos (2\theta)| \, d\theta = -\log\delta$$

This is true for all $\delta > 0$, and, therefore,

$$\int_{[0,1]^2} |f| \geqslant \lim_{\delta \to 0} (- \log \delta) = + \infty$$

Answer 4

Let $T$ be the lower triangular subregion of $[0,1]^2$ whose vertices are $(0,0),(1,0),(1,1).$ Then, letting $dA$ denote area measure, we have

$$\int_{[0,1]^2}|f|\,dA \ge \int_T |f|\,dA.$$

Time for Fubini. The integral on the right equals

$$\int_0^1\int_0^x\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx.$$

In the inner integral, let $y=xz.$ Then $dy=xdz$ and the integral becomes

$$\int_0^1\int_0^1\frac{x^2(1-z^2)}{x^4(1+z^2)^2}\,xdz\,dx = \int_0^1\frac{1}{x}\int_0^1\frac{1-z^2}{(1+z^2)^2}\,dz\,dx.$$

In the last integral, the inner integral is some positive number $C.$ We are left with

$$\int_0^1\frac{C}{x}\,dx =\infty.$$

This shows $f$ is not integrable on $[0,1]^2.$