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 $\alpha \in \mathbb{R}$
 A: $$ \begin{align} \int_{0}^{\infty} \frac{\sin (x) \sin (2x)}{x^{\alpha}} \ dx &=\frac{1}{2} \int_{0}^{\infty} \frac{\cos (x) - \cos (3x)}{x^{\alpha}} \ dx \\ &= \frac{1}{2} \int_{0}^{1} \frac{\cos (x) - \cos(3x)}{x^{\alpha}} \ dx + \frac{1}{2} \int_{1}^{\infty} \frac{\cos (x) - \cos(3x)}{x^{\alpha}} \ dx \end{align}$$
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 $0 < \alpha < 3$.
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$\ds{\int_{0}^{\infty}{\sin\pars{x}\sin\pars{2x} \over x^{\alpha}}\,\dd x:\
     {\large ?}}$.

\begin{align}&\color{#c00000}{%
\int_{0}^{\infty}{\sin\pars{x}\sin\pars{2x} \over x^{\alpha}}\,\dd x}
=2\int_{0}^{\infty}{1 \over x^{\alpha - 2}}\,
\half\int_{-1}^{1}\expo{\ic kx}\,\dd k\half\int_{-1}^{1}\expo{-2\ic qx}\,\dd q
\\[3mm]&=\half\int_{-1}^{1}\int_{-1}^{1}\
\overbrace{\int_{0}^{\infty}x^{2 - \alpha}\expo{-\pars{2q - k}\ic x}\,\dd x}
^{\ds{\mbox{Set}\ t \equiv \pars{2q - k}\ic x\ \imp\ x = {t \over k - 2q}\,\ic}}
\ \dd k\,\dd q
\\[3mm]&=\half\int_{-1}^{1}\int_{-1}^{1}
\int_{0}^{\ic\sgn\pars{2q - k}\infty}\pars{{t \over k - 2q}\,\ic}^{2 - \alpha}\expo{-t}\,{\ic\,\dd t \over k - 2q}\,\dd k\,\dd q
\\[3mm]&=\half\int_{-1}^{1}\int_{-1}^{1}\pars{\ic \over k - 2q}^{3 - \alpha}
\int_{0}^{\ic\sgn\pars{2q - k}\infty}t^{2 - \alpha}\expo{-t}\,\dd t\,\dd k\,\dd q
\\[3mm]&=\half\int_{-1}^{1}\int_{-1}^{1}\pars{\ic \over k - 2q}^{3 - \alpha}\
\overbrace{\int_{0}^{\infty}t^{2 - \alpha}\expo{-t}\,\dd t}
^{\ds{=\ \Gamma\pars{3 - \alpha}}}\
\ \dd k\,\dd q\tag{1}
\end{align}

The $\Gamma$-integral converges whenever
$\quad\ds{\Re\pars{2 - \alpha} > -1\quad\imp\quad\Re\pars{\alpha} < 3}$. In addition, we explicitly omit an integral over an arc of radius $\ds{R}$. That integral $\ds{\to 0}$, when $\ds{R \to \infty}$, as
$\ds{R^{2 - \alpha}\pars{1 - \expo{-R}}}\quad$ which requires
$\quad\ds{\Re\pars{2 - \alpha} < 0\quad\imp\quad\Re\pars{\alpha} > 2}$.
$$\color{#c00000}{\large%
\mbox{So, our result}\ \pars{1}\ \mbox{is valid whenever}\
2 < \Re\pars{\alpha} < 3}
$$

Then\,
  \begin{align}&
\color{#c00000}{%
\int_{0}^{\infty}{\sin\pars{x}\sin\pars{2x} \over x^{\alpha}}\,\dd x}
\\[3mm]&=\half\,\Gamma\pars{3 - \alpha}\int_{-1}^{1}\int_{-1}^{1}
\verts{k - 2q}^{\alpha - 3}
\exp\pars{\ic\pi\pars{3 - \alpha}\sgn\pars{k - 2q} \over 2}\,\dd k\,\dd q
\\[3mm]&=-\,\half\,\ic\Gamma\pars{3 - \alpha}\int_{-1}^{1}\int_{-1}^{1}
\verts{k - 2q}^{\alpha - 3}\sgn\pars{k - 2q}
\exp\pars{{-\pi\alpha\sgn\pars{k - 2q} \over 2}\,\ic}\,\dd k\,\dd q
\\[3mm]&=\half\,\Gamma\pars{3 - \alpha}\sin\pars{\pi\alpha \over 2}\
\underbrace{\int_{-1}^{1}\int_{-1}^{1}\verts{k - 2q}^{\alpha - 3}\,\dd k\,\dd q}
_{\ds{=\ {3^{\alpha - 1} - 1 \over \pars{\alpha - 1}\pars{\alpha - 2}}}}
\end{align}

Since
$\ds{\Gamma\pars{3 - \alpha}=\pars{2 - \alpha}\pars{1 - \alpha}
\Gamma\pars{1 - \alpha}}$:
\begin{align}&\color{#66f}{\large
\int_{0}^{\infty}{\sin\pars{x}\sin\pars{2x} \over x^{\alpha}}\,\dd x
=\half\,\pars{3^{\alpha - 1} - 1}\Gamma\pars{1 - \alpha}
\sin\pars{\pi\alpha \over 2}}
\\[3mm]&\color{#000}{\large 2 < \Re\pars{\alpha} < 3}
\end{align}
