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Consider $([-1,1]\times[-1,1],\mathcal{B}[-1,1]\times\mathcal{B}[-1,1],dx\,dy)$. Let $f(x,y)=\dfrac{xy}{(x^2+y^2)^2}$. Show the double integral does not exist.

My solution: \begin{align} \int_{[-1,1]\times[-1,1]}|f(x,y)|\,dx\,dy&\ge\int_{x^2+y^2\le1}\left|\frac{xy}{(x^2+y^2)^2}\right|\,dx\,dy\tag{1}\\ &=\int_0^{2\pi}d\theta\int_0^1\left|\frac{r\cos\theta\cdot r\sin\theta}{r^4}\right|\,r\,dr\tag{2}\\ &\ge\int_0^{\pi/2}\cos\theta\cdot\sin\theta \,d\theta\int_0^1\frac{1}{r}\,dr\tag{3}\\ &=\frac12\ln r|_0^1\tag{4}\\ &=+\infty.\tag{5} \end{align} Any comment is appreciated.

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up vote 3 down vote accepted

You immediately changed the region over which you're integrating from a square to a circle, so that won't work.

I would construe the assertion that the double integral does not exist to mean that the integrals of the positive and negative parts are both infinite. If one of them is infinite and the other is finite, I would say the integral exists and is either $+\infty$ or $-\infty$. By Fubini's theorem, the positive and negative parts must both be infinite if the two iterated integrals are unequal, i.e. $$ \int_{-1}^1\left(\int_{-1}^1 f(x,y)\,dx\right)\, dy \ne \int_{-1}^1\left(\int_{-1}^1 f(x,y)\,dy\right)\, dx. $$ The symmetry of the function and of the region over which we're integrating should cut our work in half. $$ \int_0^1 \frac{xy}{(x^2+y^2)^2}\,dx = \int_0^{\pi/4} \frac{y^2\tan u}{(y^2\tan^2 u+y^2)^2} \, \left(y\sec^2 u \, du\right) $$ $$ = \int_0^{\pi/4} \frac{y^2\tan u}{(y^2\tan^2 u+y^2)^2} \, \left(y\sec^2 u \, du\right) = \int_0^{\pi/4} \frac{y^2\tan u}{y^4\sec^2 u} \, \left(y \sec^2 u\,du\right) $$ $$ = \frac{1}{y}\int_0^{\pi/4} \tan u\,du = \left.\frac{-1}{y} \log(\cos u)\right|_0^{\pi/4} = \frac{\log 2}{2y}. $$ The integral of this function of $y$ from $0$ to $1$ is $+\infty$. So over that square we get $+\infty$, and by symmetry, over each of the second and fourth quadrants we get $-\infty$ (and over the third we get $+\infty$).

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(I've corrected several errors as this answer has evolved while I've repeatedly saved further edits, and I'm not sure I'm done yet! I wonder why no one's castigated me for the mistakes even while someone up-voted this. I wonder if they'll think better of it.........) –  Michael Hardy Dec 9 '12 at 21:02
    
I have checked that the iterated integrals are both equal to 0. –  Sam Dec 9 '12 at 21:04
    
@Sam : Did your way of checking take into account the fact that the thing blows up at $0$? Maybe what you found is a Cauchy principal value, which is $0$. –  Michael Hardy Dec 9 '12 at 21:07
    
You can't separate integrals in that way except when you have a function of $x$ times a function of $y$. –  Michael Hardy Dec 9 '12 at 21:12
    
I use Mathematica and also get 0. –  Sam Dec 9 '12 at 21:24
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There's no need to attempt to evaluate the integral. Simply consider the neighbourhood around $(0,0)$.

As $(x,y) \rightarrow (0,0)$ from the $x$- or $y$-axis, $f(x,y) \rightarrow 0$. As $(x,y) \rightarrow (0,0)$ along the line $y=x$, $f(x,y) \rightarrow \infty$.

This means that a partition around $(0,0)$ will have differing upper and lower sums for any arbitrarily small partition, so $f$ is not Riemann integrable on $[-1,1]\times[-1,1]$.

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Thank you, @mbizzle. Let me think about it. –  Sam Dec 9 '12 at 21:38
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But the mere fact that $f\to\infty$ is not enough to guarantee that the integral is infinite. Sometimes that happens and the integral is finite. For example $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}<\infty$. –  Michael Hardy Dec 9 '12 at 21:41
    
(Although in this case one probably could get by with less than evaluating the integral.) –  Michael Hardy Dec 9 '12 at 21:43
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