The problem is to evaluate the following integral (where $0<k<2, \mu >0$)
$$\int_0^{\infty} dx \, \frac{\sin^3(\mu x)}{x^k}. $$
This integral shouts to be evaluated using residues, which in turn implies to take the Principal Value because of the singularity at 0. For the contour I would choose a quarter circle in the upper right plane which goes around 0 with radius $\epsilon$. What to do next?
The result is: $$\int_0^{\infty} dx \, \frac{\sin^3(\mu x)}{x^k} = \frac{3-3^{k-1}}{4} \, \mu^{k-1} \, \Gamma(1-k) \, \cos(\frac{k\pi}{2}) .$$
$$ \begin{align} \int_0^\infty\frac{\sin^3(\mu x)}{x^k}\,\mathrm{d}x &=\frac34\int_0^\infty\frac{\sin(\mu x)}{x^k}\,\mathrm{d}x -\frac14\int_0^\infty\frac{\sin(3\mu x)}{x^k}\,\mathrm{d}x\tag{1}\\ &=\frac34\mu^{k-1}\int_0^\infty\frac{\sin(x)}{x^k}\,\mathrm{d}x -\frac14(3\mu)^{k-1}\int_0^\infty\frac{\sin(x)}{x^k}\,\mathrm{d}x\tag{2}\\ &=\frac{3-3^{k-1}}4\mu^{k-1}\int_0^\infty\frac{\sin(x)}{x^k}\,\mathrm{d}x\tag{3}\\ &=\frac{3-3^{k-1}}4\mu^{k-1}\mathrm{Im}\left(\int_0^\infty\frac{e^{ix}}{x^k}\,\mathrm{d}x\right)\tag{4}\\ &=\frac{3-3^{k-1}}4\mu^{k-1}\mathrm{Im}\left(e^{(1-k)i\pi/2}\int_0^\infty\frac{e^{-x}}{x^k}\,\mathrm{d}x\right)\tag{5}\\ &=\frac{3-3^{k-1}}4\mu^{k-1}\sin\left(\frac{(1-k)\pi}2\right)\Gamma(1-k)\tag{6}\\ &=\frac{3-3^{k-1}}4\mu^{k-1}\cos\left(\frac{k\pi}2\right)\Gamma\left(1-k\right)\tag{7} \end{align} $$ Explanation:
$(1)$: $\sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)$
$(2)$: substitute $x\mapsto\frac{x}\mu$ and $x\mapsto\frac{x}{3\mu}$
$(3)$: algebra
$(4)$: $\sin(x)=\mathrm{Im}\!\left(e^{ix}\right)$
$(5)$: use the contour $[r,R]\cup Re^{i\pi\left[0,\frac12\right]}\cup[iR,ir]\cup re^{i\pi\left[\frac12,0\right]}$ as $r\to0$ and $R\to\infty$
$\phantom{(5)\text{:}}$ noting the absence of poles inside
$\phantom{(5)\text{:}}$ the integrals along $Re^{i\pi\left[0,\frac12\right]}$ and $re^{i\pi\left[\frac12,0\right]}$ tend to $0$
$\phantom{(5)\text{:}}$ no poles inside the contour $\implies$ the integrals on $[r,R]$ and $[iR,ir]$ cancel
$\phantom{(5)\text{:}}$ thus, the integrals on $[r,R]$ and $[ir,iR]$ are equal
$(6)$: use the Gamma function and $\mathrm{Im}\!\left(e^{ix}\right)=\sin(x)$
$(7)$: $\cos(x)=\sin\!\left(\frac\pi2-x\right)$