Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx \\ &= \frac{1}{2} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{x^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{(x-i \epsilon)^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty}  \frac{x^{3}+\frac{1}{8i} (e^{3ix}-3e^{ix})}{(x-i \epsilon)^{5}} + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{-ix}-e^{-3ix}}{(x-i \epsilon)^{5}} \ dx \end{align}
Then I integrated $ f(z) = \frac{z^{3}+ \frac{1}{8i}(e^{3iz}-3e^{iz})}{(z-i \epsilon)^{5}}$ around the upper half of $|z|=R$ and $ g(z) = \frac{3e^{-iz}-e^{-3iz}}{(z-i \epsilon)^{5}}$ around the lower half of $|z|=R$ and applied Jordan's lemma. 
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}x}{x^{5}} \ dx &= \frac{1}{2} \lim_{\epsilon \to 0^{+}}2 \pi i \ \text{Res}[f(z),i \epsilon] + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} 2 \pi i (0) \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}}  \frac{2 \pi i}{4!} \lim_{z \to i \epsilon} \frac{d^{4}}{dz^{4}} \Big(z^{3}+\frac{1}{8i}e^{3iz}-\frac{3}{8i}e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \   \lim_{z \to i \epsilon}\Big( \frac{1}{8i}(3i)^{4}e^{3iz}- \frac{3}{8i} (i)^{4} e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \Big( \frac{81}{8i}e^{- 3\epsilon} - \frac{3}{8i}e^{- \epsilon} \Big) \\ &= \frac{\pi i}{24} \Big(\frac{81}{8i}-\frac{3}{8i} \Big) \\ &= \frac{13 \pi}{32} \end{align}

Here is another contour integration approach. Note that the integrand is even, so using the contours $\gamma=[-R,R]\cup Re^{i[0,\pi]}$ as $R\to\infty$ and $\beta=[-R,R]\cup Re^{-i[0,\pi]}$, we have
$$
\begin{align}
\int_0^\infty\frac{x^3-\sin^3(x)}{x^5}\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{x^3-\sin^3(x)}{x^5}\mathrm{d}x\tag{1}\\
&=\lim_{\epsilon\to0^+}\frac12\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{z^3-\sin^3(z)}{z^5}\mathrm{d}z\tag{2}\\
&=\lim_{\epsilon\to0^+}\frac12\int_{\gamma-i\epsilon}\frac{\frac1{8i}(e^{3iz}-3e^{iz})}{z^5}\mathrm{d}z\tag{3}\\
&+\lim_{\epsilon\to0^+}\frac12\int_{\beta-i\epsilon}\frac{z^3+\frac1{8i}(3e^{-iz}-e^{-3iz})}{z^5}\mathrm{d}z\tag{4}\\
&=\lim_{\epsilon\to0^+}\frac1{16i}\int_{\gamma-i\epsilon}\frac{e^{3iz}-3e^{iz}}{z^5}\mathrm{d}z\tag{5}\\[3pt]
&=\frac\pi8\frac1{4!}\left((3i)^4-3i^4\right)\tag{6}\\[6pt]
&=\frac{13\pi}{32}\tag{7}
\end{align}
$$
Explanation:
$(1)$: integrand is even, duplicate the domain and divide by $2$
$(2)$: offset the path of integration since there are no singularities and the integrand decays at $\infty$
$(3)$: take some of the terms along the upper contour
$(4)$: take the rest along the lower contour
$(5)$: there are no singularities inside $\beta-i\epsilon$ and the respective integrands decay appropriately on the circular arcs
$(6)$: the residues depend on the $z^4$ terms in the expansions of the exponentials
