Quick way to see that $\sqrt{1+x^2+y^2}^{-4}$ is $L^2$-integrable

Question

Is there a quick trick to determine that a function like $$f(x,y):=\sqrt{1+x^2+y^2}^{-4}$$ (which is essentially a negative power of the norm on $\mathbb R^d$ with the singularity at $0$ removed) is in $L^2(\mathbb R^2)$? $f$ is clearly bounded (it takes valeus in $[0,1]$), but that doesn't help here.

Answer 1

Hint $$\int_{\Bbb{R}^2} f^2=\sigma(S^1)\int_0^{\infty}\frac{r}{(\sqrt{1+r^2})^8}dr=\sigma(S^1)\int_0^{\infty}\frac{r}{(1+r^2)^4}dr$$ by polar coordinates.

Answer 2

Hint

$$ \sqrt{1+x^2+y^2}^{-4}=\frac{1}{(1+x^2+y^2)^2}=\frac{1}{1+x^2+y^2} \cdot \frac{1}{1+x^2+y^2}\leq \frac{1}{1+x^2} \cdot \frac{1}{1+y^2} $$

Answer 3

You can split the integral of $f^2$ into two pieces, $$ \int |f(x)|^2dx=\int_{|x|\le 1}|f(x)|^2\ dx+\int_{|x|>1} |f(x)|^2 \,dx $$ where $x=(x_1,x_2)\in\mathbb{R}^2$.

You have no problem for the first piece $\int_{|x|\le 1}$ since, as you observed, $f$ is bounded. For the second piece, note that $$ |f(x)|^2=\frac{1}{(1+|x|^2)^4}\le \frac{1}{|x|^8}\tag{1} $$

It is fairly easy to see by polar coordinates that $$ \int_{|x|>1} \frac{1}{|x|^8}\,dx<\infty $$