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Prove that function $f(x,y)=\dfrac{1}{x^2+y^2}$ is not Lebesgue integrable on $A=(0,1]\times(0,1]$.

To my knowledge the fastest way to do it is to use Fubini's theorem. From what I would get: $$\int_0^1 \frac{1}{x^2+y^2}dy=\frac{1}{x^3}\operatorname{arctg}\left(\frac{y}{x}\right)\Big|^1_0=\frac{1}{x^3}\operatorname{arctg}\left(\frac{1}{x}\right)$$

$$\int_0^1 \frac{1}{x^3}\operatorname{arctg}\left(\frac{1}{x}\right)\,dx=\cdots$$ Which is quite a lot of computing and I have a hard time believing that's the point of this task.

So my question is: How do I prove this function is not integrable using Fubini's theorem the fastest way possible?

EDIT. There are already two solutions posted to this question, both of which use substitution of variables but I'd like to avoid if possible, because at the point of doing this exercise we didn't know that theorem. Can anyone think any other solution (also other than just computing that complicated integral).

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up vote 10 down vote accepted

Since $\arctan x \xrightarrow[x\to\infty]{}\frac{\pi}{2}$, $\frac{1}{x^3}\arctan\frac{1}{x}\operatorname*{\sim}_{x\to 0}\frac{\pi}{2x^3}$. But since $x\mapsto\frac{1}{x^3}$ is not Lebesgue integrable on a neighborhood of $0$, so isn't $x\mapsto\frac{1}{x^3}\arctan\frac{1}{x}$: $$ \int_{[0,1]}\frac{1}{x^3}\arctan\frac{1}{x}\mu(dx) = +\infty $$

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That's exactly the answer I was looking for! – Max Feb 3 '14 at 0:40

We shall show that $$ \int_C f(x,y)\,dx\,dy=\infty, $$ where $$ C=\{(x,y): x,y\ge 0,\,\,\, \text{and}\,\,\, x^2+y^2\le 1\}. $$ Using polar coordinates, i.e., $$ x=r\cos\vartheta,\,\, y=r\sin\vartheta\quad\text{and}\quad dx\,dy=r\,dr\,d\vartheta,$$ we obtain $$ C=\big\{(r,\vartheta): r\in[0,1],\,\,\vartheta\in[0,2\pi]\big\}, $$ and thus $$ \int_C \frac{1}{x^2+y^2}\,dx\,dy=\int_0^{\pi/2}\left(\int_0^1 \frac{1}{r^2}r\,dr\right)\,d\vartheta=\frac{\pi}{2}\int_0^1\frac{dr}{r}=\infty. $$

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@Max You could edit your question and add that you're after a solution that does not use a change of variable to clarify this. – Geoff Pointer Feb 3 '14 at 0:31
@Geoff Pointer - well that's a good point... – Max Feb 3 '14 at 0:36

Hint: use polar coordinates in the neighbourhood of zero.

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