= Solution
{c}
Near zero, $\sin x/x\to1$, so $|\sin x|/x^\alpha$ is comparable to $x^{1-\alpha}$. The <power integral test> shows that the integral near zero is finite exactly when $\alpha<2$.
For the tail, $|\sin x|\leq1$ proves integrability when $\alpha>1$. To prove necessity, take the disjoint <intervals>
$$
I_k=[2k\pi+\pi/6,\ 2k\pi+5\pi/6],\qquad k\geq1.
$$
On each, $\sin x\geq1/2$, and $x\leq2\pi(k+1)$. Therefore $\int_{I_k}|\sin x|x^{-\alpha}\,dx\geq c_\alpha(k+1)^{-\alpha}$ with $c_\alpha>0$. The <p-series> diverges when $\alpha\leq1$, proving that the absolute tail integral diverges in that range.
Combining the two conditions,
$$
\boxed{f_\alpha\text{ is Lebesgue integrable on }(0,\infty)\iff1<\alpha<2.}
$$
For $0<\alpha\leq1$, the tail <improper integral> converges conditionally by the <Dirichlet test for improper integrals>, but that does not give <Lebesgue integrability>.
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