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. :(

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 $\alpha > 0$
