I am studying residue calculations and until now had learned that one should avoid using it for integral of rational functions P/Q when Q has real roots. However, I found an exercise concerning this integral :
$$\int_{-\infty}^{\infty}\frac{1}{(x+1)(x^2-x+1)}dx $$
which obviously has a real roots for $x=-1$… The answer I saw was to include one half the residue for pole $Z=-1$ when going complex. I never heard of that : could anyone point to any source and explain why it is so ?
Thanks and have a good day !
Marc
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{r \equiv }$: \begin{align} &\bbox[5px,#ffd]{\mrm{P.V.}\int_{-\infty}^{\infty} {\dd x \over \pars{x + 1}\pars{x^{2} - x + 1}}} \\[5mm] = &\ \overbrace{2\pi\ic\lim_{\large x\ \to\ \expo{\ic\pi/3}}\,\,\, {x - \expo{\ic\pi/3} \over \pars{x + 1}\pars{x^{2} - x + 1}}} ^{\ds{{2 \over 3}\,\pi\,\ic\expo{-2\pi\ic/3}}} \\[2mm] - &\ \underbrace{\lim_{\large \epsilon \to 0^{+}}\,\,\int_{\pi}^{0} {\epsilon\expo{\ic\theta}\ic\,\,\dd\theta \over \epsilon\expo{\ic\theta} \bracks{\pars{-1}^{2} - \pars{-1} + 1}}}_{\ds{-\,{\pi \ic \over 3}}} \\[5mm] = &\ \bbx{{\root{3} \over 3}\,\pi} \approx 1.8138 \\ & \end{align}