Prove integral diverges

Question

How can I prove that the improper integral: $\int_0^\infty x^\alpha\sin (x) \,dx$ diverges for $\alpha>0$?

I can clearly integrate by parts to reduce the exponent on $\alpha$, but then I get a lot of limits tending to infinity.

Can I claim the integral diverges since all of them tend to $+\infty$?

Edit: In fact the limits don't all tend to $+\infty$, they tend to $-\infty$ as well, so that doesn't seem to work...

Answer 1

You may show by elementary inequalities that both the sequences $\{a_k\}_{k\geq 1}$ and $\{b_k\}_{k\geq 1}$ are divergent, where $$ a_k=\int_{0}^{(2k-1)\pi}x^\alpha\sin(x)\,dx,\qquad b_k=\int_{0}^{2k\pi}x^\alpha\sin(x)\,dx. $$ For instance $$\int_{k\pi}^{(k+1)\pi}x^{\alpha}\left|\sin x\right|\,dx \sim 2(\pi k)^\alpha $$ follows from the mean value theorems for integrals.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{R > 0\,,\ \alpha \in \mathbb{R}\ \mbox{and}\ z^{\alpha} = \verts{z}^{\alpha}\exp\pars{\ic\arg\pars{z}\alpha}\,,\ z \not= 0\,,\ -\pi < \arg\pars{z} < \pi}$:

\begin{align} \int_{0}^{R}x^{\alpha}\sin\pars{x}\,\dd x & = \Im\int_{0}^{R}x^{\alpha}\expo{\ic x}\dd x \,\,\,\stackrel{x\ =\ \ic t}{=}\,\,\, \Im\int_{0}^{-R\ic}\pars{\ic t}^{\alpha}\expo{-t}\ic\,\dd t \\[5mm] & = -\,\Re\int_{-\pi/2}^{0}\pars{\ic R\expo{\ic\theta}}^{\alpha} \expo{-R\expo{\ic\theta}}R\expo{\ic\theta}\ic\,\dd\theta - \Re\int_{R}^{0}t^{\alpha}\expo{\ic\pi\alpha/2}\expo{-t}\,\dd t \\[5mm] & = -\,\Re\int_{-\pi/2}^{0}\pars{\ic R\expo{\ic\theta}}^{\alpha} \expo{-R\expo{\ic\theta}}R\expo{\ic\theta}\ic\,\dd\theta + \cos\pars{{\pi \over 2}\,\alpha} \color{#00f}{\int_{0}^{R}t^{\alpha}\expo{-t}\,\dd t}\label{0}\tag{0} \end{align}

The $\color{#00f}{\mbox{second integral}}$ converges whenever $\ds{\alpha > -1}$.

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\begin{align} 0 & < \verts{\Re\int_{-\pi/2}^{0}\pars{\ic R\expo{\ic\theta}}^{\alpha} \expo{-R\expo{\ic\theta}}R\expo{\ic\theta}\ic\,\dd\theta} \\[5mm] & < R^{\alpha + 1}\int_{0}^{\pi/2}\expo{-R\cos\pars{\theta}}\dd\theta = R^{\alpha + 1}\int_{0}^{\pi/2}\expo{-R\sin\pars{\theta}}\dd\theta \\[5mm] & < R^{\alpha + 1}\int_{0}^{\pi/2}\expo{-2R\theta/\pi}\dd\theta = R^{\alpha + 1}\,{\expo{-R} - 1 \over -2R/\pi} = {1 \over 2}\,\pi\pars{R^{\alpha} - R^{\alpha}\expo{-R}} \,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\, \color{red}{{1 \over 2}\,\pi R^{\alpha}} \label{1}\tag{1} \end{align}

$\color{red}{\mbox{Expression}}$ \eqref{1} $\ds{\to 0}$ whenever $\ds{\alpha < 0}$.

Finally, $$ \bbx{\left.\int_{0}^{\infty}x^{\alpha}\sin\pars{x}\,\dd x \,\right\vert_{\ \alpha\ \in\ \mathbb{R}} = \cos\pars{{\pi \over 2}\,\alpha}\Gamma\pars{\alpha + 1}\,,\qquad \alpha \in \pars{-1,0}} $$