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Background: Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be defined as $$f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$$ and integrate $f$ over the unit square $U=[0,1]\times[0,1]$, show that $$\iint_U |f|\text{ }dA=\infty$$

Attempt: Since the set $B=\{(r\cos\theta,r\sin\theta)\mid r\in[0,1],\theta\in[0,\pi/2] \}$ is contained in $U$, we should have $\iint_B|f|dA\leq\iint_U|f|dA.$ But \begin{align} (*)\dots\iint_B|f|dA=\int_0^{\pi/2}\int_0^1 \frac{|\cos^2\theta-\sin^2\theta|}{r}drd\theta=\int_0^{\pi/2}|\cos^2\theta-\sin^2\theta|d\theta\cdot\int_0^1r^{-1}dr \end{align} Observe that $\int_0^1 r^{-1}dr$ diverges, and $\int_0^{\pi/2}|\cos^2\theta-\sin^2\theta|d\theta$ converges, the proof is completed.

But clearly, we cannot have (at least based on the Fubini's theorem) the first equality in $(*)$ since $r^{-1}$ is not integrable on $[0,1]$ (i.e., if the hypothesis of the Fubini's theorem is not satisfied). Do I have a way to make the above proof correct?

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    $\begingroup$ What if you argue by contradiction? Suppose the integral is convergent. Then, by Fubini's theorem, such and such would hold, but we arrive at a contradiction Finite = Infinite. $\endgroup$ – GReyes Mar 17 '19 at 18:22
  • $\begingroup$ You had the right idea. $\endgroup$ – RRL Mar 17 '19 at 20:22
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There's a counterpart to Fubini's theorem known as Tonelli's theorem: if $g$ is nonnegative everywhere, then $\int_a^b\int_c^d g(x,y)\,dx,\,dy = \int_c^d\int_a^b g(x,y)\,dy\,dx$. This includes the case that if one of them is infinite, so is the other. As such, we're free to write $\iint_U g(x,y)\,dA$, and evaluate it in any form we wish.

So then, the integral of the absolute value you're trying to calculate there is independent of order. You've calculated directly that it's infinite in one way, and thus $\iint_U |f|\,dA = \infty$.

The way to use Fubini's theorem isn't calculating the integral of $|f|$. Instead, we calculate the iterated integrals $\int_0^1\int_0^1 f(x,y)\,dx\,dy$ and $\int_0^1\int_0^1 f(x,y)\,dy\,dx$: \begin{align*}\int_0^1\int_0^1 f(x,y)\,dx\,dy &= \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy\\ &\phantom{|}^{x=y\tan\theta}_{dx=y\sec^2\theta\,d\theta}\\ &= \int_0^1\int_0^{\arctan(1/y)}\frac{y^2(\tan^2\theta-1)}{(y^2\sec^2\theta)^2}\cdot y\sec^2\theta\,d\theta\,dy\\ &= \int_0^1\int_0^{\arctan(1/y)}\frac{-\cos 2\theta}{y}\,d\theta\,dy\\ &= \int_0^1 -\frac{\sin(2\arctan(1/y))}{2y}\,dy = \int_0^1\frac{-1}{1+y^2}\,dy = -\frac{\pi}{4}\\ \int_0^1\int_0^1 f(x,y)\,dy\,dx &= \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx\\ &\phantom{|}^{y=x\tan\theta}_{dy=x\sec^2\theta\,d\theta}\\ &= \int_0^1\int_0^{\arctan(1/x)}\frac{x^2(1-\tan^2\theta)}{(x^2\sec^2\theta)^2}\cdot x\sec^2\theta\,d\theta\,dx\\ &= \int_0^1\int_0^{\arctan(1/x)}\frac{\cos 2\theta}{x}\,d\theta\,dx\\ &= \int_0^1 \frac{\sin(2\arctan(1/y))}{2y}\,dy = \int_0^1\frac{1}{1+y^2}\,dy = \frac{\pi}{4}\end{align*} The two iterated integrals are not equal. Therefore, the conditions of Fubini's theorem are not satisfied. The only condition that could fail is absolute integrability, so therefore $\iint_U |f(x,y)|\,dA = \infty$.

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With some care, you can use Fubini's theorem as you started — to show that $\int_U |f| = +\infty$. We first excise a small sector containing the origin so that we are dealing with proper Riemann integrals and then take the limit as the radius of the sector tends to $0$.

Take $D_\delta = \{ (r,\theta): 0 < r < \delta, 0 \leqslant \theta \leqslant \pi/2 \}$ and $D = \{ (r,\theta): \delta \leqslant r \leqslant 1, 0 \leqslant \theta \leqslant \pi/2 \}$. Since $|f|$ is nonnegative and $D\setminus D_\delta \subset U\setminus D_\delta$, we have

$$\int_{U\setminus D_\delta} |f| \geqslant \int_{D\setminus D_\delta} |f| $$

Since $|f|$ is bounded and continuous on $D \setminus D_\delta$ we can compute the double integral as the iterated integral in any order

$$\int_{D\setminus D_\delta} |f| = \int_0^{\pi/2} \int_\delta^1|f| \,r \, dr \, d\theta = \int_0^{\pi/2}|\cos^2\theta-\sin^2\theta| \, d\theta\cdot\int_\delta^1r^{-1}\, dr \\= -\log \delta \int_0^{\pi/2}|\cos^2\theta-\sin^2\theta|\, d\theta$$

Thus, the improper integral diverges since

$$\int_{U} |f| = \lim_{\delta \to 0}\int_{U\setminus D_\delta} |f| \geqslant -\lim_{\delta \to 0} \log \delta \int_0^{\pi/2}|\cos^2\theta-\sin^2\theta| \, d\theta = +\infty$$

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