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}

Another approach :
Sorry Random Variable, this is not using contour integration technique since I don't know how to approach the integral using that way. $\ddot\smile$
Consider
$$
\mathcal{I}(\alpha)=\int_0^\infty\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}dx.\tag1
$$
Differentiating $(1)$ four times yields
\begin{align}
\frac{d^4\mathcal{I}}{d\alpha^4}&=\int_0^\infty\frac{\partial^4}{\partial\alpha^4}\left(\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}\right)dx\\
&=\color{green}{\int_0^\infty\left(\frac{81\sin3\alpha x-3\sin\alpha x}{4x}\right)dx}\\
&=\frac{81}{4}\cdot\frac\pi2-\frac{3}{4}\cdot\frac\pi2\\
&=\frac{39\pi}{4},\tag2
\end{align}
where
$$
\int_0^\infty\frac{\sin\alpha x}{x}dx=\frac\pi2\qquad\text{for }\alpha\neq0.
$$
Then from $(2)$ we obtain
$$
\large\color{blue}{\mathcal{I}(\alpha)=\frac{13\pi}{32}a^4},\tag3
$$
where $\mathcal{I}(0)=\mathcal{I'}(0)=\mathcal{I''}(0)=\mathcal{I'''}(0)=0$. Thus

$$
\mathcal{I}(1)=\int_0^\infty\frac{x^3-\sin^3 x}{x^5}dx=\large\color{blue}{\frac{13\pi}{32}}.
$$


P.S.
I think you can easily apply the contour integration technique in line $2$ (green-colored) equation $(2)$.
