Show $\int_0^\infty x^{-\alpha}\sin x dx$ exists for $\alpha \in (0,2)$. 
Show that $\int_0^\infty x^{-\alpha}\sin x dx$ exists for $\alpha \in (0,2)$.

This is a real analysis class question.  I am not quite sure how to show this.  I tried a whole bunch of things like integration by parts, bounding sine, ... and nothing worked. Any hints, ideas, or solutions would be appreciated.
 A: Consider separately 
$$\int_0^1 x^{-\alpha}\sin x\, dx \qquad\text{and}\qquad \int_1^\infty x^{-\alpha}\sin x\, dx.$$
For the first integral, use the fact that for $0<x\le 1$ we have $0\le \sin x \le x$.  Now prove the convergence of the first integral by comparing our function with $\dfrac{1}{x^{\alpha-1}}$, noting that $\alpha-1<1$.
For the second integral, let
$$I(M)=\int_1^M x^{-\alpha} \sin x\,dx$$
and examine the behaviour of $I(M)$ as $M\to\infty$.
There are various ways to proceed. One is to integrate by parts, letting $u=x^{-\alpha}$ and $dv=\sin x\,dx$.  We get something which behaves nicely as $M$ gets large, and an integral of something that is a constant times $\dfrac{\cos x}{x^{1+\alpha}}$.  This integral behaves nicely as $M\to\infty$, by comparison of $\dfrac{|\cos x|}{x^{1+\alpha}}$ with $\dfrac{1}{x^{1+\alpha}}$.
The reason for the integration by parts was to increase the power of $x$ at the bottom, in order to make the function decrease faster, fast enough for obvious convergence of the integral.  
A: Kb100 has exactly the right idea. We can split the proof into three parts:


*

*Show that $\int_0^\epsilon x^{-\alpha} \sin x dx$ exists only if $\alpha<2$ by utilizing $\sin x = x +\mathcal{O}(x^3)$.

*Observe that the improper integral diverges for $\alpha\le0$ because $\lim\limits_{x\to\infty} x^{-\alpha}\sin x \ne 0$.

*For $\alpha\in(0,2)$ split the integral into an alternating series (note $\sin$ is nonnegative on $[0,\pi]$), $$\int_0^\pi x^{-\alpha}\sin xdx - \int_0^\pi (x+\pi)^{-\alpha}\sin x dx+\int_0^\pi (x+2\pi)^{-\alpha}\sin x dx -\cdots $$ Prove that the terms above are decreasing in magnitude by first proving that $x^{-\alpha}>(x+\pi)^{-\alpha}$ for all $x>0$ and positive $\alpha\;$ (hint: $(1+\epsilon)^\alpha>1$). Now just invoke the alternating series test.
A: For the integral $\int_1^\infty x^{-\alpha}\sin x\, dx$ we may apply Dirichlet's Convergence Test for Improper Integrals, witch states:
Let


*

*function $f(x)$ is integrable in closed interval $[a,A]$ $(A>a)$, and 


$$|\int_a^Af(x)dx|\leq K \ \ \ \ \ \  (K=\text{const},a\leq A<\infty)$$.


*

*function $g(x)$ is monotone and 


$$\lim_{x\to\infty}g(x)=0.$$
Then $\int_a^{\infty}f(x)g(x)dx$ converges.

In our case, 
$$ |\int_a^A\sin(x)dx|=|\cos(a)-\cos(A)|\leq 2 $$,
and, $\frac{1}{x^{\alpha}}$ is monotonic decreasing to $0$.
