convergence of a generalized Riemann integral Could you please provide me some hints to test the convergence of this integral below?
$$\int\limits_{0}^{+\infty}\dfrac{\sin x\cdot \sin 2x}{x^\alpha} \, dx$$ where $\alpha \in \mathbb{R}$

$$ \begin{align} \int_{0}^{\infty} \frac{\sin (x) \sin (2x)}{x^{\alpha}} \ dx &=\frac{1}{2} \int_{0}^{\infty} \frac{\cos (x) - \cos (3x)}{x^{\alpha}} \ dx \\ &= \frac{1}{2} \int_{0}^{1} \frac{\cos (x) - \cos(3x)}{x^{\alpha}} \ dx + \frac{1}{2} \int_{1}^{\infty} \frac{\cos (x) - \cos(3x)}{x^{\alpha}} \ dx \end{align}$$
Expanding $\cos (x)$ and $\cos (3x)$ in Maclaurin series,  you can see that $\cos(x) - \cos(3x)$ behaves like $4x^{2}$ near $x=0$.
So the first integral converges if $2-\alpha > -1$.  That is, if $\alpha <3$.
And 
$$ \begin{align}  \int_{1}^{b} \Big( \cos (x) - \cos (3x) \Big) \ dx &= \sin(x) - \frac{1}{3} \sin(3x) \Bigg|_{1}^{b} \\  &= \frac{4}{3} \sin^{3}(x) \Big|^{b}_{1} \\ &= \frac{4}{3} \Big( \sin^{3}(b) - \sin^{3}(1)\Big) \end{align}$$
which remains bounded for any value of $b$ greater than $1$.
So by Dirichlet's convergence test, the second integral converges if $\alpha >0$.
See here for more information about the test.
Therefore, the original integral converges if $0 < \alpha < 3$.
