when does $\int_{0}^{\infty}\frac{\sin(t)}{(t+1)^\alpha}dt$ converge?

Question

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)

Answer 1

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

Answer 2

For $\alpha \lt 1$, $\int_0^\infty \frac{sin(t)}{(1+t)^\alpha}dt=\sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\frac{sin(t)}{(1+t)^\alpha}dt$. The terms are decreasing in magnitude. The even $n$ terms are positive and the odd $n$ terms are negative, so the series and therefore the integral is convergent, but not absolutely convergent.

I looked at the comment about alternating series and it is valid. Combing series terms in pairs will give absolute convergence for $0\lt \alpha \le 1$. We need to collapse adjacent terms of the sum and get $\sum_{n=0}^\infty \int_{2n\pi}^{(2n+1)\pi}sin(t)(\frac{1}{(1+t)^\alpha}-\frac{1}{(1+t+\pi)^\alpha})$. In the interval $[2n\pi,(2n+1)\pi]$, $t=2n\pi+x$, where $0\le x \le \pi$. Therefore $\frac{1}{(1+t)^\alpha}-\frac{1}{(1+t+\pi)^\alpha}=\frac{1}{(1+t)^\alpha}(1-\frac{1}{(1+u_n)^\alpha})$, where $u_n=\frac{\pi}{1+2n\pi+x}\approx\frac{1}{2n}$, for large $n$, and $\frac{1}{(1+u_n)^\alpha}\approx 1-\frac{\alpha}{2n}$ Also $\frac{1}{(1+t^\alpha)}\approx \frac{1}{(2n\pi)^\alpha}$. Combining these estimates and get $\frac{\alpha}{\pi^\alpha (2n)^{\alpha+1}}$ for large $n$. Since $\alpha+1\gt 1$, the series with collapsed terms converges.