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...

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.
