for which a values $\int_{\mathbb{R^{2}}} \frac{sin(x^2+y^2)}{(x^2+y^2+1)^a}dxdy $ converges?

Question

For which a values $\displaystyle \int_{\mathbb{R^{2}}} \frac{\sin(x^2+y^2)}{(x^2+y^2+1)^a}\,dx\,dy$ converge? I believe that only for $a\geq 0.5$ but I don't how to prove it. I was able to reduce the problem for only one variable $\displaystyle \frac{1}{2r^2+1}$ but I don't know how to prove this converges for $a>0.5$ or diverges for $a\leq 0.5$.

Answer 1

Transforming to polar coordinates we find that

$$\int_{\mathbb{R}^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^a}\,dx\,dy=2\pi \int_0^\infty\frac{\sin(\rho^2)}{(\rho^2+1)^a}\,\rho\,d\rho$$

Then, enforcing the substitution $\rho\mapsto \sqrt u$ reveals $$\int_{\mathbb{R}^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^a}\,dx\,dy=\pi \int_0^\infty \frac{\sin(u)}{(u+1)^a}\,du\tag 1$$ whereby the Abel-Dirichlet test guarantees that the integral in $(1)$ converges for all $a>0$.