Integrating with Fubini's theorem

Question

I have to find $\alpha, \beta \in \Bbb R$ for which $$ \int_{\Bbb R^2} \frac{dx}{|x|_{\infty}^\alpha(1+|x|_{\infty}^\beta)} < \infty$$ using Fubini's theorem. Note that $x = (x_{1},x_{2})$ and $|x|_{\infty}=\max(|x_{1}|, |x_{2}|)$ .

Any help/tip would be greatly appreciated!

Answer 1

$\|\cdot\|_1,\|\cdot \|_2$ and $\|\cdot\|_{\infty}$ are topologically equivalent, hence the answer is unchanged if we consider $$ \iint_{\mathbb{R}^2}\frac{dx\,dy}{(x^2+y^2)^{\alpha/2}(1+(x^2+y^2)^{\beta/2})}=2\pi\int_{0}^{+\infty}\frac{\rho}{\rho^{\alpha}(1+\rho^{\beta})}\,d\rho$$ which is convergent as soon as $\alpha<2$ and $\alpha+\beta >2$.
In exact terms $$ \iint_{\mathbb{R}^2}\frac{d\mu}{\|x\|_\infty^\alpha(1+\|x\|_\infty^\beta)} =8\int_{0}^{+\infty}\frac{r}{r^\alpha(1+r^\beta)}\,dr=\frac{8\pi}{\beta\sin\frac{\pi(2-\alpha)}{\beta}}.$$

Answer 2

You can split the plane into 3 regions according to whether $|x_1|$ is less, more, or equal to $|x_2|$. The last one has measure zero.

Study the integral on the first region by first integrating with respect to $x_1$ and then estimating the integral wrt $x_2$. Here you use Fubini.

The second region gives the same value by symmetry.

This way you will find that the integral is finite if and only if $\alpha<2$ and $\alpha+\beta>2$.