Integral converges/diverges Question:
$$p \in \Bbb R  \int _0 ^\infty \frac {\sin x}{x^p}dx$$
For what values of p does this integral converge/diverge?
Thoughts
We've tried using Dirichlet criteria and bounding it from below, after splitting it from $[0,1]$ and $[1.\infty]$ There seems to be some kind of trick we are probably missing.
 A: Observe that the oscillatory property of the sine function accounts for some amount of cancellation, yielding better convergent behavior.
One way of removing this effect is to turn the oscillatory effect into a better decreasing rate by exploiting integration by parts:
$$ \int_{1}^{R} \frac{\sin x}{x^{p}} \, dx = \left[ \frac{1-\cos x}{x^{p}} \right]_{1}^{R} + p \int_{1}^{R} \frac{1-\cos x}{x^{p+1}} \, dx. $$
Now the right-hand side exhibits no oscillatory behavior due to the non-negativeness of $1-\cos x$. This shows that the integral converges as $R \to \infty$ if and only if $p > 0$.
For the near-zero behavior of the integrand, just approximate $\sin x$ by polynomials using Taylor series: $\sin x = x + O(x^{3})$. Then
$$ \int_{\epsilon}^{1} \frac{\sin x}{x^{p}} \, dx = \int_{\epsilon}^{1} \frac{dx}{x^{p-1}} + \int_{\epsilon}^{1} O(x^{3-p}) \, dx, $$
which converges as $\epsilon \to 0$ if and only if $p < 2$.
Thus the near-infinity behavior and the near-zero behavior tie to yield a convergent integral exactly for $0 < p < 2$.
A: For $p = 1$, it is convergent. You can show it by using.
$$
\int_{1}^{b}\frac{\text{sin}\;x}{x}dx = -\frac{\text{cos}\; b}{b} + \text{cos}\;1 - \int_{1}^{b}\frac{\text{cos}\;x}{x^2}dx
$$
The point is, you can show by using integral by parts. 
A: The sharp border is $p \geq 1$. for $p = 1$ you have
$${\rm sinc}(x) = \frac{\sin x}x$$
About which you can show
$$\int_a^b {\rm sinc}(x)\ dx \leq C < \infty \qquad \forall a,b \in \mathbb R$$
