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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.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{x - \sin\pars{x} \over x^{3}\pars{1 + x^{2}}}\,\dd x}= {1 \over 2}\,\Im\int_{-\infty}^{\infty}{\ic x - \expo{\ic x} + 1 - x^{2}/2 \over x^{3}\pars{1 + x^{2}}}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Im\braces{2\pi\ic\,% {\ic\pars{\ic} - \expo{\ic\pars{\ic}} + 1 - \ic^{2}/2 \over \ic^{3}\pars{\ic + \ic}}} = \bbx{\large{\expo{} -2 \over 4\expo{}}\,\pi} \\ & \end{align}


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}
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  • $\begingroup$ 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. $\endgroup$ – 2132123 Aug 15 at 4:01
  • $\begingroup$ @2132123 The main purpose is to avoid the divergence of the integral under $\displaystyle\Im$ without modifyng, of course, its value. Thanks. $\endgroup$ – Felix Marin Aug 15 at 4:08
  • $\begingroup$ Can some one explain or just name the theory/method behind this amazing solution. A good ref./siurce will hrlp me more. Please respond to help me I have never seen such a simple solution of a complicated integral. $\endgroup$ – Dharmendra Singh Aug 15 at 7:54
  • $\begingroup$ @DharmendraSingh this is just residue thoerem. Felix sorry to bother you again, but how do you justify the convergence of the integral? $\endgroup$ – 2132123 Aug 15 at 15:37
  • $\begingroup$ @2132123 I add something about that. Please, check the answer very end. $\endgroup$ – Felix Marin Aug 16 at 14:21

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