Showing the integral $\int_{\mathbb{R}} \int_{\mathbb{R}} \min\{ 1, (\max \{ |x|,|y| \})^{-3} \} dx dy$ converges

Question

I am trying to bound the following integral: $\int_{\mathbb{R}} \int_{\mathbb{R}} \min\{ 1, (\max \{ |x|,|y| \})^{-3} \} dx dy$. I am very sure this integral converges, but whatever I try seem to get messy very quickly. Is there a non-very messy approach for this? Thank you very much!

Answer 1

Let $ f(x,y)=\min\{ 1, (\max \{ |x|,|y| \})^{-3} \}\leq 1$. Then $$ |x|\geq 1\vee |y|\geq 1\Rightarrow f(x,y)=\frac{1}{[\max\{|x|,|y|\}]^3}\leq\left[\frac{1}{(|x|+|y|)/2}\right]^3\leq\left[\frac{2}{\sqrt{|x|^2+|y|^2}}\right]^3=8r^{-3} $$ So $f$ is bounded by 1 and out of the square $Q=\{(x,y)\,|\,|x|<1,|y|<1\}$ is bounded by radial integrable funtion $r^{-3}$.

Answer 2

For $r^2 = x^2 + y^2 > 1$, you have $\max( |x|, |y| ) \ge \frac{\sqrt{2}}{2} r$, because a square fits inside a circle. :)

So split your integral into the the part over the unit disk (clearly finite, since the integrand is bounded) and the part outside the unit disk. Call that $I$. Then $$ I \le \int_0^{2\pi} \int_{r=1}^\infty k^3 r^{-3} r ~ dr ~ d\theta $$ where $k = 2 / \sqrt{2}$ is a constant that can be ignored. The inner integrand ($r^{-2}$) anti-differentiates to $r^{-1}/(-1)$, and evaluates to 1. The outer integral throws a factor of $2\pi$. So your integral $I$ is no more than $2\pi k^3$, hence bounded.

Answer 3

It is not difficult to compute the integral exactly, by symmetry.

Let $D=\{(x,y)\in\mathbb{R}^+\times\mathbb{R}^+: y\leq x\}$. Then:

$$\begin{eqnarray*} I = 8\iint_{D}\min\left(1,\frac{1}{x^3}\right)\,dx\,dy &=& 8\int_{0}^{+\infty}\int_{0}^{x}\min\left(1,\frac{1}{x^3}\right)\,dy\,dx\\&=&8\int_{0}^{+\infty}\min\left(x,\frac{1}{x^2}\right)\,dx\\&=&8\int_{0}^{1}x\,dx + 8\int_{1}^{+\infty}\frac{dx}{x^2}=\color{red}{12}.\end{eqnarray*}$$