when does $\int_{0}^{\infty}\frac{\sin(t)}{(t+1)^\alpha}dt$ converge? The original question was to determine for which values of $\alpha \in \mathbb R$ does the integral $$\iint_{\mathbb R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^\alpha}dxdy$$ converge.
I managed to simplify this and even reach a partial answer:
$$\iint_{\mathbb R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^\alpha}dxdy = 2\pi\int_{0}^{\infty}r\frac{\sin(r^2)}{(r^2+1)^\alpha}dr = \pi\int_{0}^{\infty}\frac{\sin(t)}{(t+1)^\alpha}dt$$
Fine. Let's investigate $\int_{0}^{\infty}\frac{\sin(t)}{(t+1)^\alpha}dt$.
$$\left|\int_{0}^{\infty}\frac{\sin(t)}{(t+1)^\alpha}dt\right| \leq \int_{0}^{\infty}\frac{1}{(t+1)^\alpha}dt = \begin{cases}\infty, \text{        }\alpha \leq 1\\-\frac{1}{1-a}, \text{          }\alpha > 1\end{cases}$$
So when $\alpha > 1$ the integral converges. Great. But we don't know what happens when $\alpha \leq 1$.
I've failed to come up with more helpful comparison tests, and the integral itself is not very pleasant, I'm sure the teacher did not intend the students to actually calculate the anti-derivative (this question was from an exam)

Hint (for the main question, not the original inspiring question): Try integration by parts with $u = (1+t)^{-\alpha}$, $dv = \sin t\,dt$.
