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

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
  $\ds{\alpha > -1}$.


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