convergence of a generalized Riemann integral

Question

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 $\alpha \in \mathbb{R}$

Answer 1

$$ \begin{align} \int_{0}^{\infty} \frac{\sin (x) \sin (2x)}{x^{\alpha}} \ dx &=\frac{1}{2} \int_{0}^{\infty} \frac{\cos (x) - \cos (3x)}{x^{\alpha}} \ dx \\ &= \frac{1}{2} \int_{0}^{1} \frac{\cos (x) - \cos(3x)}{x^{\alpha}} \ dx + \frac{1}{2} \int_{1}^{\infty} \frac{\cos (x) - \cos(3x)}{x^{\alpha}} \ dx \end{align}$$

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 $0 < \alpha < 3$.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\sin\pars{x}\sin\pars{2x} \over x^{\alpha}}\,\dd x:\ {\large ?}}$.

\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}