TITLE: Find the values $a>0$ for which this improper integral converges
QUESTION [6 upvotes]: Find the values of $a > 0$ for which the improper integral $$\int_{0}^{\infty} \frac{\sin x} {x^a} dx$$ converges.

This question is from my analysis quiz (now over).
I have studied improper integrals but Dirichlet test, Comparison tests and other results can't be used in this case as Dirichlet test is used if $\sin x$ used to tend to $\infty$, also 2 comparison tests are not suited for two functions inside the integral.
If I use Abel test and write
$$\int_{0}^{\infty} \frac{\sin x} {x^a} dx = \int_{0}^{a} \frac{\sin x} {x^a} dx + \int_{a}^{\infty} \frac{\sin x} {x^a} dx$$
then $\int_{a}^{\infty} \frac{\sin x} {x^a} dx$ converges at infinity for $a<1$ but  $\sin x$ is not bounded and monotonic in $(a, \infty)$. So, it can't be used as well.
So, what result should I use. (I have done a course on real analysis.)

REPLY [3 votes]: Fact #1: The improper integral $\int_{\pi}^{\infty}\frac{\sin x}{x^a}\mathrm{d}x$ converges for all $a>0$
Proof of Fact #1: Let $t>\pi$. From integration by parts, $$\int_\pi^t\frac{\sin x}{x^a}\mathrm{d}x=-\frac{\cos t}{t^a}-\frac{1}{\pi^a}-a \int_{\pi}^{t}\frac{\cos x}{x^{a+1}}\mathrm{d}x$$ Since $a>0$ we know $\frac{\cos t}{t^a}\rightarrow 0$ as $t \rightarrow \infty$ by squeeze theorem. Moreover $a+1>1$ and $\bigg|\frac{\cos x}{x^{a+1}}\bigg|\leq \frac{1}{x^{a+1}}$ for any $x>0$ so $\lim_{t \rightarrow \infty}\int_{\pi}^{t}\frac{\cos x}{x^{a+1}}\mathrm{d}x$ exists. In fact, $\int_{\pi}^{\infty}\frac{\cos x}{x^{a+1}}\mathrm{d}x$ converges absolutely. This show our integral converges for $a>0$.
Fact #2: The improper integral $\int_{0}^{\pi}\frac{\sin x}{x^a}\mathrm{d}x$ converges if and only if $a<2$.
Proof of Fact #2: Since $\frac{\sin x}{x}\rightarrow 1$ as $x\rightarrow 0^{+}$ there is $0<\delta < \pi$ so that $\Bigg|\frac{\sin x}{x}-1\Bigg|<\frac{1}{2}$ for $0<x<\delta$. This means $\frac{1}{2x^{a-1}}<\frac{\sin x}{x^a}<\frac{3}{2x^{a-1}}$ for $x\in (0,\delta)$. Because $\int_{0}^{\delta}\frac{\mathrm{d}x}{x^{a-1}}$ converges if and only if $a<2$ we get with direct comparison test $\int_0^{\delta}\frac{\sin x}{x^a}\mathrm{d}x$ converges if and only if $a<2$. The result follows by noting $\int_\delta^{\pi}\frac{\sin x}{x^a}\mathrm{d}x$ isn't even improper.
Combine Fact #1 and Fact #2 to see how $\int_0^{\infty}\frac{\sin x}{x^a}\mathrm{d}x$ converges if and only if $a<2$.