Does the improper integral $\int\limits_0^{+\infty}x^p\sin x\,\mathrm{d}x,~~p>0$ converge?

Question

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.

Answer 1

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.

Answer 2

\begin{align*} &\int_{0}^{M}x^{p}\sin xdx\\ &=-M^{p}\cos M+p\int_{0}^{M}x^{p-1}\cos xdx\\ &=-M^{p}\cos M+p\int_{0}^{1}x^{p-1}\cos xdx\\ &~~~~~~~~+pM^{p-1}\sin M-p\sin 1-p(p-1)\int_{1}^{M}x^{p-2}\sin xdx. \end{align*} The term $\displaystyle\int_{1}^{\infty}x^{p-2}\sin xdx$ is absolutely convergent for $0<p<1$. But the term $-M^{p}\cos M+pM^{p-1}\sin M$ has no limit as $M\rightarrow\infty$.

Answer 3

I will repeat (not verbatim) an argument another user made here, which for some reason he deleted but I don't know why (it seems correct to me). There might be some mistake in this argument, and if so feel free to let me know.

There exists an absolute constant $C>0$ and infinitely many disjoint compact intervals $\{I_n\}_{n \geq 1}$ of length $C$ in $(1, \infty)$ such that $\sin(x) > \frac{1}{2}$ whenever $x \in \bigcup_{n} I_n$. Now put $g(x) = \int_{0}^{x} t^p \sin t \ dt$ where $p>0$ is arbitrary. If $\lim_{x \to \infty} g(x)$ exists, then $\lim_{n \to \infty} g(\sup I_n) - g(\inf I_n) = 0$, but this is absurd since $g(\sup I_n) - g(\inf I_n) \geq \frac{C}{2}$ for each $n$.

Answer 4

If you can use the Dirichlet's test, then there is an easy argument: suppose that $\displaystyle{\int^{+\infty}_{0}} x^p\sin x{\rm d}x$ converges, then since $\dfrac{1}{x^p}$ monotonally decreases to $0$ on $[1,+\infty)$, the integral $\displaystyle{\int^{+\infty}_{1}} \sin x{\rm d}x$ must also converge, but it does not.