I'm working through a contour integral question, which is rounded off by finding the integral:
$\int^{\infty}_{0} \frac{x-\sin(x)}{x^3} dx$
I have already shown that the residue at $0$ of the function
$f(z)=\frac{1+iz-e^{iz}}{z^3}$
on $\mathbb{C} - \{0\}$ is $\frac{1}{2}$, and that
$\int_{\gamma_R} f(z) dz \longrightarrow 0$ as $R \longrightarrow 0$ where $\gamma_R:[0,\pi]\rightarrow \mathbb{C}$ is given by $\gamma_R(t)=Re^{it}$.
Problem
How do I progress from here to finding the required integral? It's an odd function, so clearly all that is left is to integrate from $R$ to $-R$ along the real axis and half it to find the integral, but I can't see how to pull out the '$x-\sin(x)$' and replace '$z^3$' with '$x^3$'.
An explanation of how to finish this off would be much appreciated. Thanks in advance.