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$$(2-x)(x^2+4)=0$$ Roots: $$(2,2i,-2i)$$
$$Res(\frac{1}{(2-x) (x^2+4)},2)=\lim_{x\rightarrow 2} \frac{1}{-(x^2+4)}=-\frac{1}{8}$$
$$Res(\frac{1}{(2-x) (x^2+4)},2i)=\lim_{x\rightarrow 2i} \frac{1}{(2-x)(x+2i)}=\frac{1}{8} \ \frac{1}{i+1} $$

$$P.V.\int_{-\infty}^{+\infty} \frac{1}{(2-x) (x^2+4)}=\frac{\pi}{4} i \frac{1}{i+1} - \frac{\pi}{8} i=\frac{\pi}{8} i (-i+1)-\frac{\pi}{8} i=\frac{\pi}{8}$$


Is it correct?


Thanks!

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    P.V. stands for Cauchy Principal Value? Also, what is this $Res$ symbol? Whatever method you're appling, I'm not sure you're allowed to do complex calculus in real analysis. What course is it for? – 2017-01-25
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    I've just noticed that the function you want to integrate over $(- \infty, + \infty)$ is not defined at $x=2$, which means that it doesn't exist, neither does the principal value! – 2017-01-25
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    (-half the residue at the real line+the residue in the upper half plane)*2 pi i – 2017-01-25

1 Answers 1

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{P.V.}\int_{-\infty}^{\infty}{\dd x \over \pars{2 - x}\pars{x^{2} + 4}} \\[5mm] \stackrel{\mbox{def.}}{=}\,\,\,&\ -\lim_{\epsilon \to 0^{+}}\bracks{% \int_{-\infty}^{2 - \epsilon}{\dd x \over \pars{x - 2}\pars{x^{2} + 4}} + \int_{2 + \epsilon}^{\infty}{\dd x \over \pars{x - 2}\pars{x^{2} + 4}}} \\[5mm] = &\ -\lim_{\epsilon \to 0^{+}} \bracks{\oint_{\mc{C}}{\dd z \over \pars{z - 2}\pars{z^{2} + 4}} - \int_{\pi}^{0}{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}\times 8}} \\[5mm] = &\ -\lim_{\epsilon \to 0^{+}}\bracks{2\pi\ic\,{1 \over \pars{2\ic - 2}\pars{4\ic}} + {1 \over 8}\,\pi\ic} = \bbx{\ds{{1 \over 8}\,\pi}} \end{align}

$\mc{C}$ 'closes' the integration in the upper complex plane with and additional indent around $\ds{z = 2}$. There is a simple pole, inside $\ds{\,\mc{C}}$, at $\ds{z = 2\ic}$.