Prove that function is not Lebesgue integrable

Question

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).

Answer 1

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 $x\mapsto\frac{1}{x^3}\arctan\frac{1}{x}$: $$ \int_{[0,1]}\frac{1}{x^3}\arctan\frac{1}{x}\mu(dx) = +\infty $$

Answer 2

Hint: use polar coordinates in the neighbourhood of zero.

Answer 3

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 $$ \int_C \frac{1}{x^2+y^2}\,dx\,dy=\int_0^{\pi/2}\left(\int_0^1 \frac{1}{r^2}r\,dr\right)\,d\vartheta=\frac{\pi}{2}\int_0^1\frac{dr}{r}=\infty. $$