How to see the convergence/divergence of this integral

Question

I have a question, I want to know for which values of $p\in \mathbb{R}$ this integral have finite value: $$\int_{0}^{\infty} \frac{\sin{x}}{x^p}dx $$ I have shown that for $p=1$ the integral is finite and equals to $\frac{\pi}{2}$, so, if $p\geq1$ is easy to see that the integral converges. The intuition says that if $p<1$, it diverges. But I don't know how to prove that. Any idea?

Answer 1

The integral converges for $0<p<2$.

Convergence for $0 < p \leq 1$. First note that $\lim_{x\to 0}(\sin x)/x^p=\lim_{x\to 0}x^{1-p} =0$ if $p < 1$; the limit is 1 if $p=1$. So there is no problem around $x=0$. Therefore it suffices to show that the integral from $1$ to $\infty$ converges.

Partial integration gives $$ \begin{align*} \int_1^\infty \frac{\sin x}{x^p}\,dx &= \int_1^\infty \frac{1}{x^{p}}\,d(-\cos x) \\ &= -\Biggl.\frac{\cos x}{x}\Biggr|_1^\infty - p\int_1^\infty \frac{\cos x}{x^{p+1}}\,dx \\ &= \cos 1 -p\int_1^\infty \frac{\cos x}{x^{p+1}}\,dx. \end{align*} $$ Since $|x^{-(p+1)}\cos x|\leq |x^{-(p+1)}|$ and since $p>0$ the integral in the last term converges. Therefore the whole integral $$ \int_0^\infty \frac{\sin x}{x^p}\,dx $$ is convergent.

Convergence for $1 < p < 2$. Note that $\sin x \sim x$ for $x\to 0$. Therefore the integrand is asymptotically equivalent with $\frac{1}{x^{p-1}}$ around $x=0$. In order for the integral to be convergent, $p-1 < 1$, or $p < 2$.

Note that at infinity there is no problem. If you would take the integral from, say, 1 to infinity, convergence follows from the inequality $|x^{-p}\sin x|\leq |x^{-p}|$ and the fact that $p > 1$.

No convergence for $p < 0$. We write $$ \int_{\pi}^\infty \frac{\sin x}{x^p}\,dx = \sum_{n=1}^\infty a_n \quad \text{ with } \quad a_n = \int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x^p}\,dx. $$ For $n\pi \leq x \leq (n+1)\pi$, we have $(n\pi)^{-p}\leq x^{-p}\leq((n+1)\pi)^{-p}$ since $-p > 0$. For even integers $n$, the $a_n$ are positive and we have $$ \int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x^p}\,dx \geq (n\pi)^{-p}\int_{n\pi}^{(n+1)\pi} \sin x\,dx = 2(n\pi)^{-p}. $$ For odd $n$, the term $a_n$ is negative and $$ \int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x^p}\,dx \leq ((n+1)\pi)^{-p}\int_{n\pi}^{(n+1)\pi} \sin x\,dx = -2((n+1)\pi)^{-p}. $$ This implies that the series $\sum_{n=1}^\infty a_n$ diverges, and so the integral also diverges.