For which values of $\alpha$ the integral $\int_{R^2} \frac{\sin(x^2+y^2)}{(x^2+y^2+1)^{\alpha}}$ converges?

Question

For which values of $\alpha$ the integral $\int_{R^2} \frac{\sin(x^2+y^2)}{(x^2+y^2+1)^{\alpha}}$ converges?

I changed to polar coordinates and got to $\int_{R^2} \frac{r\sin(r^2)}{(r^2+1)^{\alpha}}$

I managed to show that for $\alpha \le0$ the integral doesn't converge and for $\alpha \ge 1$ it converges.

However, for $0 < \alpha < 1$ , I am not sure how to bound it.

Help would be appreciated.

Answer 1

After switching to polar coordinates the integral becomes

$$ 2\pi\int_{0}^{+\infty}\frac{\rho\sin(\rho^2)}{(\rho^2+1)^{\alpha}}d\rho=\pi\int_{0}^{+\infty}\frac{\sin(x)}{(x+1)^{\alpha}}\,dx $$ which converges for any $\alpha>0$ due to Dirichlet's criterion: $\sin(x)$ has a bounded antiderivative and $\frac{1}{(x+1)^{\alpha}}$ decreases to zero.