Does the improper integral $\int\limits_0^{+\infty}x^p\sin x\,\mathrm{d}x$ for $p>0$ converge?
Attempt. Limit of $x^p\sin x$ as $x$ tends to $+\infty$ does not exist, in order to guarantee integral's divergence. So I worked on the definition. For $p=1$ we get the integral:
$$\int\limits_0^{+\infty}x\sin x\,\mathrm{d}x=\lim_{x\to +\infty}(\sin x-x\cos x)$$ (after integration by parts), which does not converge (the limit does not exist). This approach also works for $p>1$, but I had difficulties regarding the case $p<1$.
Thank you in advance.
Best Answer
If that integral converged, then as $n\to \infty$ through integer values,
$$\int_{2\pi n}^{2\pi n + \pi} x^p\sin x\, dx\to 0.$$
But this integral is greater than
$$(2\pi n)^p\int_{2\pi n}^{2\pi n + \pi} \sin x\, dx = (2\pi n)^p\cdot 2 \to \infty,$$ contradiction.