Find the values $a>0$ for which this improper integral converges 
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.)
 A: 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 $a<2$.
A: I'll admit that I'm not very good at proofs, but I believe I've got something; I have reason to believe the following:
$$ \int_0^\infty \dfrac{\sin(x)}{x^{1+\frac{1}{\alpha}}} \textrm{d}x = \alpha \sin\left(\dfrac{\pi}{2\alpha}\right)\Gamma\left(\dfrac{\alpha-1}{\alpha}\right) $$
I believe this holds for all $\alpha\in(1,\infty)$, or rather $a\in[1,2)$ if you wanna revert to the original integral. I'll leave some updates later on if additional information is required.
Disclaimer:
Just for clarification, I am establishing that this identity holds for all $a\in[1,2)$, not necessarily that the integral in question converges for only $a\in[1,2)$. So far, this is only the beginning, and I plan to add later details in the future in case I stumble across any more values of $a$ for which the integral converges.

Edit 1: I forgot to specify that the following integral DEFINITELY converges for all $a\in(0,2)$:
$$ \int_0^\infty \dfrac{\sin(x)}{x^a} \textrm{d}x = \dfrac{\cos\left(\frac{a\pi}{2}\right)\Gamma(2-a)}{1-a} $$
The reason why the interval only runs from $0$ to $2$ is because, when $a=2$, the RHS is equal to $\dfrac{\cos\left(\frac{2\pi}{2}\right)\Gamma(2-2)}{1-2}$ or $\Gamma(0)$, which we know blows up to infinity. Conversely, when $a=0$, the RHS is equal to $1$, which can't happen because $\int_0^\infty \sin(x) \textrm{d}x$ doesn't converge.
A: First let us split the integral into two parts:
$$
\int_{0}^{\infty} \frac{\sin x} {x^a} \, dx
= \int_{0}^{\pi} \frac{\sin x} {x^a} \, dx
+ \int_{\pi}^{\infty} \frac{\sin x} {x^a} \, dx
.
$$
The first integral,
$$
\int_{0}^{\pi} \frac{\sin x} {x^a} \, dx
= \int_{0}^{\pi} \frac{x+O(x^3)} {x^a} \, dx
= \int_{0}^{\pi} (x^{1-a}+O(x^{3-a})) \, dx
$$
is convergent if $1-a>-1,$ i.e. if $a<2.$
The second integral,
$$
\int_{\pi}^{\infty} \frac{\sin x} {x^a} \, dx
= \sum_{k=1}^{\infty} \int_{k\pi}^{(k+1)\pi} \frac{\sin x} {x^a} \, dx
$$
is an alternating series where the terms decrease in absolute value to zero monotonically for every $a>0$ so it is convergent for all $a>0$ by the alternating series test.
Thus, the original integral converges if $a<2.$
