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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.

Onur
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Sam
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2 Answers2

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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 $+\infty$).

Michael Hardy
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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 09 '12 at 21:02
  • I have checked that the iterated integrals are both equal to 0. – Sam Dec 09 '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 09 '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 09 '12 at 21:12
  • I use Mathematica and also get 0. – Sam Dec 09 '12 at 21:24
  • Probably Mathematica is giving you a Cauchy principal value. If $\int_x^1 g(w),dw\to+\infty$ as $x\downarrow0$ and $\int_{-1}^x g(w),dw\to-\infty$ as $x\uparrow0$, then the Cauchy principal value of $\int_{-1}^1 g(w),dw$ is $\lim_{x\downarrow0}\left( \int_{-1}^{-x} + \int_x^1 \right) g(w) , dw$. – Michael Hardy Dec 09 '12 at 21:36
  • Thank you very much, @Michael. – Sam Dec 09 '12 at 21:49
  • @MichaelHardy I know this is very late but I had a question about something you wrote above and I've spent a few hours building up to 50 reputation so I can add a comment! You said: "If one of them is infinite and the other is finite, I would say the integral exists and is either positive or negative infinity". If one of the parts is infinite though, that means that the integral of the absolute value of the function is also infinite. Which means that Fubini's theorem doesn't apply. And yet, the double integral exists. Is Fubini's theorem failing independent of a double integral existing? – Blahblahblacksheep Mar 26 '20 at 18:32
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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 $[-1,1]\times[-1,1]$.

mbarkley
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  • Thank you, @mbizzle. Let me think about it. – Sam Dec 09 '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 09 '12 at 21:41
  • (Although in this case one probably could get by with less than evaluating the integral.) – Michael Hardy Dec 09 '12 at 21:43