For what values of $\alpha$ does the following improper integral converges?
$$\int_{\Bbb R^2} \frac{\sin(x^2+y^2)}{(1+x^2+y^2)^\alpha}dxdy$$
I tried to change to polar coordinates and look at the absolute value of the integral. It seems like the answer should be $\alpha >1$ but I'm not sure how to prove it.
Call your integral $I(\alpha)$. Indeed, polar coordinates allows us to write this as: $$\begin{align}I(\alpha)&=\int_0^{2\pi}\int_0^\infty \frac{\sin(r^2)}{(1+r^2)^\alpha}r~drd\theta\\&=\pi\int_0^\infty\frac{\sin(u)}{(1+u)^\alpha}~du\qquad u=r^2\implies du=2r~dr\end{align}$$ Can you find for what $\alpha$ the integral $$J(\alpha):=\int_0^\infty\frac{\sin(u)}{(1+u)^\alpha}~du$$ converges?