Cauchy integral indented contour question: $\int_{-\infty}^\infty \frac{(x+1)^2}{x(x^2+4)^2}dx$

Question

I need to calculate the followig integral using residue calculus with an indented contour: $$\int_{-\infty}^\infty \frac{(x+1)^2}{x(x^2+4)^2}dx$$

It would be great if someone could show me how to do it but one thing in particular that I'm unsure about is what to do about the integral concerning the semi-circular contour around the origin. If $\epsilon$ is the radius of the semi-circular contour at the origin, should the limit as $\epsilon \rightarrow 0$ of the integral vanish (or not), and in any case what do I need to do with the integral and how do I do it?

Answer 1

On a little semicircle $\;C_\epsilon\;$ (upper) of radius $\;\epsilon>0\;$ about zero, you indeed have to take the limit $\;\epsilon\to0\;$ and then you get by the corollary to the lemma here , that

$$\lim_{\epsilon\to0}\int_{C_\epsilon}\frac{(z+1)^2}{z(z^2+4)^2}dz=\pi i\,\text{Res}\,(f)_{z=0}=\frac{\pi i}{16}$$

The rest is standard: the upper semicircle of radius $\;R>>0\;$ so that you only need the poles with positive imaginary part, take residues, limits and etc.

Added on request: The actual "problem" is to deal with the integral on the arc $\;\gamma_R:=\{z=Re^{it}\;/\;0<t<\pi\}\;$ , but either the L-M (Estimmation Lemma) or Jordan's Lemma solve that and this part of the integral, in this case, tends to zero when $\;R\to\infty\;$.

Also, the residue at $\;z=2i\;$, which is a double pole (observe that we don't care about the other pole $\;z=-2i\;$, as we are on the upper semiplane) is

$$\text{res}\,(f)_{z=2i}=\lim_{z\to 2i}\left((z-2i)^2\frac{(z+1)^2}{z(z^2+4)^2}\right)'=\lim_{z\to2i}\left(\frac{(z+1)^2}{z(z+2i)^2}\right)'$$$${}$$

$$=\lim_{z\to2i}\frac{2z(z+1)(z+2i)-(z+1)^2\left((z+2i)+2z\right)}{z^2(z+2i)^3}=\frac{4i(1+2i)\cdot4i-(1+2i)^2(4i+4i)}{(-4)(-64i)}=$$$${}$$

$$=\frac{-16-32i-(-3+4i)\cdot8i}{256i}=\frac{-16-32i+24i+32}{256i}=-\frac1{32}-\frac i{16}$$

so we get, calling $\;C\;$ the whole simple, closed contour:

$$-\frac1{32}-\frac i{16}=\oint_{C}f(z)dz=\int_{-R}^{-\epsilon}f(x)dx-\int_{\gamma_\epsilon}f(z)dz+\int_\epsilon^Rf(x)dx+\int_{\gamma_R}f(z)dz$$

Observe the minus sign before the integral around zero because when we go from $\;-R\to R\;$ that half semicircle's integration is done in the negative direction! Well, now just take the double limit and use Cauchy's Theorem:

$$\lim_{R\to\infty,\,\epsilon\to0}\left(-\frac1{32}-\frac i{16}\right)=\frac1{2\pi i}\lim_{R\to\infty,\,\epsilon\to0}\int_Cf(z)dz=\frac1{2\pi i}\left(\int_{-\infty}^\infty f(x)dx-\frac{\pi i}{16}\right)\implies$$$${}$$

$$\implies\int_{-\infty}^\infty\frac{(x+1)^2}{x(x^2+4)^2}dx=\frac\pi8 $$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Assuming you are interested in the Integral Principal Value $\ds{\pars{~\mathrm{P.V.}~}}$:

\begin{align} \color{#f00}{\mrm{P.V.}\int_{-\infty}^{\infty}{\pars{x + 1}^{2} \over x\pars{x^{2} + 4}^{2}}\,\dd x} & = \int_{0}^{\infty}{\pars{x + 1}^{2} - \pars{-x + 1}^{2} \over x\pars{x^{2} + 4}^{2}}\,\dd x = 4\int_{0}^{\infty}{\dd x \over \pars{x^{2} + 4}^{2}}\,\dd x \\[5mm] & = \left.-4\,\partiald{}{\mu}\int_{0}^{\infty}{\dd x \over x^{2} + \mu} \,\dd x\,\right\vert_{\ \mu\ =\ 4} = -4\,\partiald{}{\mu}\bracks{\mu^{-1/2}\int_{0}^{\infty}{\dd x \over x^{2} + 1} \,\dd x}_{\ \mu\ =\ 4} \\[5mm] & = -4\,\bracks{-\,\half\,\mu^{-3/2}\,\,\pars{\pi \over 2}}_{\, \mu\ =\ 4} = \color{#f00}{\pi \over 8} \end{align}