How to show that $\dfrac{\sin(x^2+y^2)}{(x^2 + y^2)^\alpha}$ integrable on $\mathbb{R}^2$

Question

I need to show that

$$k(x,y) = \dfrac{\sin(x^2+y^2)}{(x^2 + y^2)^\alpha}$$ is integrable on $\mathbb{R}^2$ for $1<\alpha <2. $

How do I go about this? I'm pretty sure I need to use Tonelli's Theorem and Fubini's Theorem, but I really am not sure how to apply them.

I've tried using the bound $k(x,y) \leq \dfrac{1}{(x^2+y^2)^\alpha}$ but this doesn't seem to be integrable.

Answer 1

By using polar coordinates: $$ \iint_{x^2+y^2\leq R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2)^{\alpha}}=2\pi\int_{0}^{R}\frac{\sin(\rho^2)}{\rho^{2\alpha-1}}\,d\rho = \pi\int_{0}^{R^2}\frac{\sin(z)}{z^\alpha}\,dz$$ and by bounding $|\sin z\,|$ with $\min(1,|z|)$ we have that $\frac{\sin z}{z^{\alpha}}\in L^1(\mathbb{R}^+)$, provided that $\alpha\in(1,2)$.