I know how to solve and prove that
$\int_{0}^{\infty} \frac{\sin(x)}{x^\alpha} \, dx$ converge for $ 0 < \alpha < 2 $ with regular tests and integration by parts.
But with the Dirichlet test, I just see that I have one function which is monotonic decreasing to $0$ for any given $\alpha$ and the integral on the $\sin(x)$ is bounded for $ [0,\infty]$ because so it should converge. so it should converge for any $\alpha$.
Clearly I don't understand the Dirichlet test, but I've read its definition many times and still I don't understand where I got it wrong.