Calculate $\int_{0}^{\infty} \frac{x-\sin(x)}{x^3(1+x^2)}$

Calculate $$\int_{0}^{\infty} \frac{x-\sin(x)}{x^3(1+x^2)}$$

I know i am supposed to use residue theorem. However, I am having trouble with the pole at $$z=0$$ normally i would try the funciton $$f(z)=\frac{z-e^{iz}}{z^3(1+z^2)}$$ but this clearly is not working as this function has a pole of order 3 at $$z=0$$. if i try to reduce the order of the pole i would need to do something along those lines: $$f(z)=\frac{z-ie^{iz}+i}{z^3(1+z^2)}$$ and here the pole is simple, but integrating this function will not give me the desired integral, I don' think. What do I do?

Edit:

Perhaps $$y=x^2$$ substitution and keyhole integration would work. I will have to check.

The $$\ds{\large\left. contribution\ from\ the\ arc\ R\expo{\ic\pars{0,\pi}}\,\right\vert_{\ R\ >\ 1}}$$:
\begin{align} 0 & < \verts{\int_{0}^{\pi} {\ic R\expo{\ic\theta} - \expo{\ic R\cos\pars{\theta}}\expo{-R\sin\pars{\theta}} + 1 - R^{2}\expo{2\ic\theta} \over R^{3}\expo{3\ic\theta}\pars{1 + R^{2}\expo{2\ic\theta}}}\,R\expo{\ic\theta}\ic\,\dd\theta} \\[5mm] & < \int_{0}^{\pi}{R + \expo{-R\sin\pars{\theta}} + 1 + R^{2} \over R^{3}\pars{R^{2} - 1}}\,R\,\dd\theta \\[5mm] & = {R^{2} + R + 1 \over R^{2}\pars{R^{2} - 1}}\,\pi + {1 \over R^{2}\pars{R^{2} - 1}} \int_{-\pi/2}^{\pi/2}\expo{-R\cos\pars{\theta}}\dd\theta \\[5mm] & \stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\, {2 \over R^{4}} \int_{0}^{\pi/2}\expo{-R\sin\pars{\theta}}\dd\theta < {2 \over R^{4}} \int_{0}^{\pi/2}\expo{-2R\theta/\pi}\dd\theta \\[5mm] & = \pi\,{1 - \expo{-R} \over R^{5}} \,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\LARGE\to}\,\,\, {\large\color{red}{0}} \end{align}
• Ugh, I thought i learned you trick by trying the $+i$ but i guess it can be taken even further :P. Thank you again for illuminating me. – 2132123 Aug 15 at 4:01
• @2132123 The main purpose is to avoid the divergence of the integral under $\displaystyle\Im$ without modifyng, of course, its value. Thanks. – Felix Marin Aug 15 at 4:08