Help Finding the Cauchy Principle Value of $\int_{0}^{2\pi}\frac{d\theta}{1+2cos(\theta)}$

Question

$$\int_{0}^{2\pi}\frac{d\theta}{1+2cos(\theta)}$$

My attempt:

parametrise using $z=e^{i\theta}$ (i think we always use a unit circle for CPV's)

$\therefore dz = ie^{i\theta}d\theta$ $\implies d\theta=\frac{dz}{iz}$

and $2cos(\theta) = e^{i\theta}+e^{-i\theta} = z+z^{-1}$

we can then re-write the integral as follows:

$$\int_{0}^{2\pi}\frac{d\theta}{1+2cos(\theta)} = \displaystyle\oint_{|z|=1}\frac{1}{(1+z+z^{-1})}\frac{dz}{iz}$$

$$\displaystyle\oint_{|z|=1}\frac{1}{(1+z+z^{-1})}\frac{dz}{iz} = \frac{1}{i}\displaystyle\oint_{|z|=1}\frac{dz}{(z^{2}+z+1)}$$

poles are at $z^2+z+1=0$ and know that $z^2+z+1=\frac{z^3-1}{z-1}$

therefore we require that $z\ne 1$ and $z^3=1$ and we know that $z=e^{i\theta}$

$\therefore z^3=e^{\pm2ki\pi}=1 \implies z = e^{\pm 2ki\pi/3}$

now the integral can be rewritten as:

$$\frac{1}{i}\displaystyle\oint_{|z|=1}\frac{dz}{(z-e^{2ki\pi/3})(z-e^{-2ki\pi/3})}$$

i think CPV = $2\pi i\sum Residues\space at\space the\space poles$=$$2\pi i\frac{1}{i} \left( \frac{1}{e^{2ki\pi/3}-e^{-2ki\pi/3}}+\frac{1}{e^{-2ki\pi/3}-e^{2ki\pi/3}}\right) = 2\pi \left( \frac{1-1}{e^{2ki\pi/3}-e^{-2ki\pi/3}}\right)=0$$

Is what i have done here correct, is this the right way the Cauchy integral theorem to find the CPV. Thanks for any help you can offer me.

Answer 1

Your method is correct, but just to make sure there are no questions, I am adding a diagram of the contour and description of the method I used.

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The Cauchy Principal Value of $$ \int_0^{2\pi}\frac{\mathrm{d}\theta}{1+2\cos(\theta)} $$ is the contour integral $$ \int_{\gamma^-}\frac1{1+z+z^{-1}}\frac{\mathrm{d}z}{iz} =\frac1i\int_{\gamma^-}\frac{\mathrm{d}z}{z^2+z+1} $$ where $\gamma$ is the closed contour

and $\gamma^-$ is $\gamma$ minus the two small arcs around the singularities at $e^{2\pi i/3}$ and $e^{4\pi i/3}$ as the radius of the arcs goes to $0$.

There are no singularities inside $\gamma$. Thus, the integral along $\gamma$ is $0$. This means that the integral along $\gamma^-$ is $\pi i$ times the sum of the residues at $e^{2\pi i/3}$ and $e^{4\pi i/3}$ (each of the small arcs is $\frac12$ a circle clockwise and we want to subtract them).

That is $$ \begin{align} \mathrm{PV}\int_0^{2\pi}\frac{\mathrm{d}\theta}{1+2\cos(\theta)} &=\pi i\left[\operatorname*{Res}_{z=e^{2\pi i/3}}\left(\frac1i\frac1{z^2+z+1}\right)+\operatorname*{Res}_{z=e^{4\pi i/3}}\left(\frac1i\frac1{z^2+z+1}\right)\right]\\ &=\pi i\left[\frac{-1}{\sqrt3}+\frac1{\sqrt3}\right]\\[6pt] &=0 \end{align} $$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{\,{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \pp\int_{0}^{2\pi}{\dd\theta \over 1 + 2\cos\pars{\theta}}} =2\pp\int_{0}^{\pi}{\dd\theta \over 1 - 2\cos\pars{\theta}} \\[5mm]&=2\pp\int_{0}^{\infty} {2\,\dd t/\pars{1 + t^{2}}\over 1 - 2\pars{1 - t^{2}}/\pars{1 + t^{2}}} =4\pp\int_{0}^{\infty}{\dd t \over 3t^{2} - 1} ={4 \root{3} \over 3}\pp\int_{0}^{\infty}{\dd t \over t^{2} - 1} \\[5mm]&={4 \root{3} \over 3}\,\lim_{\epsilon\ \to\ 0^{+}} \pars{\int_{0}^{1 - \epsilon}{\dd t \over t^{2} - 1} +\int_{1 + \epsilon}^{\infty}{\dd t \over t^{2} - 1}} \\[5mm]&={4 \root{3} \over 3}\,\lim_{\epsilon\ \to\ 0^{+}} \pars{\int_{0}^{1 - \epsilon}{\dd t \over t^{2} - 1} +\int_{1/\pars{1 + \epsilon}}^{0}{-\dd t/t^{2} \over 1/t^{2} - 1}} \\[5mm]&={4 \root{3} \over 3}\,\lim_{\epsilon\ \to\ 0^{+}} \pars{-\int_{0}^{1 - \epsilon}{\dd t \over 1 - t^{2}} +\int_{0}^{1/\pars{1 + \epsilon}}{\dd t \over 1 - t^{2}}} \\[5mm]&={4 \root{3} \over 3}\,\lim_{\epsilon\ \to\ 0^{+}} \dsc{\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}{\dd t \over 1 - t^{2}}} =\color{#66f}{\LARGE 0} \end{align}

because

\begin{align}& 0<\verts{\dsc{\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}{\dd t \over 1 - t^{2}}}} <\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}} {\dd t \over 1 - 1/\pars{1 + \epsilon}^{2}} ={\pars{1 + \epsilon}^{2} \over \epsilon^{2} + 2\epsilon} \bracks{{1 \over 1 + \epsilon} - \pars{1 - \epsilon}} \\[5mm]&={\pars{1 + \epsilon}^{2} \over \pars{\epsilon + 2}\pars{\epsilon + 1}}\, \epsilon\quad\to\quad 0\quad\mbox{when}\quad\epsilon \to 0 \end{align}