Convergence of $\int_{1}^{\infty} \frac{\sin x}{x^{\alpha}}dx$ For which values of $\alpha > 0$ is the following improper integral convergent?
$$\int_{1}^{\infty} \frac{\sin x}{x^{\alpha}}dx$$
I tried to solve this problem by parts method but I am nowhere near to the answer. :(
 A: $\frac{1}{x^\alpha}$ is a continuous decreasing function on $[1,+\infty)$ converging to zero and $\sin x$ is a function with a bounded primitive, hence for any $\alpha >0$ the integral
$$ \int_{1}^{+\infty}\frac{\sin x}{x^\alpha}\,dx$$
is converging due to the integral version of Dirichlet's test. Integration by parts leads to:
$$\int_{1}^{M}\frac{\sin x}{x^\alpha} = \left.\frac{-\cos x}{x^\alpha}\right|_{1}^{M}-\int_{1}^{M}\frac{\alpha\cos x}{x^{\alpha+1}}\,dx=\cos 1+O\left(\frac{1}{M^\alpha}\right)+O(1).$$
A: As David Mitra mentioned in the comments you can split the integral up. First we do 
$$\int_{1}^{\infty} \frac{\sin x}{x^{\alpha}}dx = \int_{1}^{\pi} \frac{\sin x}{x^{\alpha}}dx + \int_{\pi}^{\infty} \frac{\sin x}{x^{\alpha}}dx$$
Now the first part converges for sure (you can approxiamate it with $\frac{1}{x^a}$). For the second 
$$ \int_{\pi}^{\infty} \frac{\sin x}{x^{\alpha}} dx = \sum_{j=1}^{\infty} \int_{j\pi}^{(j+1)\pi} \frac{\sin x}{x^{\alpha}} dx \le \sum_{j=1}^{\infty} \int_{j\pi}^{(j+1)\pi} \frac{\sin x}{(j\pi)^{\alpha}} dx =
\sum_{j=1}^{\infty} \frac{1}{(j\pi)^{\alpha}} \int_{j\pi}^{(j+1)\pi} \sin x dx = \sum_{j=1}^{\infty} 2 (-1)^j \frac{1}{(j\pi)^{\alpha}} $$ 
and since $\frac{2}{(j\pi)^{\alpha}} \to 0$ for $j \to \infty$ and $(-1)^j$ is alternating, the series converges (and so the integral)  for every $\alpha > 0$
