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Given $a > b > 0$ what is the fastest possible way to evaluate the following integral using Residue theorem. I'm confused weather to take the imaginary part of $z^2$ or whole integral. $$\int_0^{2 \pi} {\cos^2 \theta \over a + b \cos \theta}\; d\theta$$

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up vote 3 down vote accepted

If $z=e^{i\theta}$, then $\cos\theta=\dfrac 1 2\left(z+\dfrac1z\right)$, and $dz=ie^{i\theta}\,d\theta/2=iz\,d\theta/2$, so $-2i\dfrac{dz}{z} = d\theta$. Then $$ \int_0^{2\pi} \frac{\cos^2\theta}{a+b\cos\theta} d\theta = \int\limits_\text{circle} \frac{\frac14\left(z+\frac1z\right)^2}{a+\frac b2\left(z+\frac1z\right)}(-i)\frac{dz}{z} = -i\int\limits_\text{circle} \frac{z^4+2z^2+1}{2z^2(2az+b(z^2+1))} dz. $$

This function has a double pole at $z=0$ and simple poles at $\dfrac{-a\pm\sqrt{a^2-b^2}}{b}$. So the question is: for which values of $a,b$ are the simple poles inside the circle? If there's just one simple pole inside the circle at $c$, then the integral becomes $$ \int\limits_\text{circle} \frac{g(z)}{z-c} dz = 2\pi i g(c), $$ where $g(z)$ is whatever's left after you've factored out $1/(z-c)$. If there's more than one simple pole, you need a sum: take values of $g$ at those points and sum them.

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Actually I found I could substitute $ \cos^2 \theta = Im(z^2)$ and simplify things a lot :D the whole integral will be $ Im ( \int z^2/...)$ – hasExams Sep 1 '12 at 17:41
We have $e^{2i\theta}=\cos[2\theta]+i\sin[2\theta]$. You take the real part ($\cos^{2}[\theta]-\sin^{2}[\theta]$), not the imaginary part (which is $2\sin[\theta]\cos[\theta]$ instead). – Bombyx mori Sep 2 '12 at 6:17
@Michael: We have $\cos[\theta]=\frac{1}{2}(\cos[\theta]+i\sin[\theta])+(\cos[\theta]-i\sin[\theta‌​])$, you missed a factor. – Bombyx mori Sep 2 '12 at 6:27
@user32240 : Fixed now, I hope. – Michael Hardy Sep 2 '12 at 18:34

Hint: put $z={\rm e}^{i\theta}$ and change the integral to the form $ \int_{|z|=1}f(z)\,dz \,,$ then use residue theorem. Exploit the identity $$ \cos\theta = \frac{1}{2}\left({\rm e}^{i \theta} + {\rm e}^{- i \theta}\right)\,. $$

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Are you suggesting doing an integral of the form $\int_{|z|=1}\frac{z^{2}}{a+bz}$? But though it has an obvious pole at $z=-\frac{a}{b}$(which is outside of the contour), I do not see how to get the original integral from $\cos[\theta]=\frac{z+\overline{z}}{2}$. But thanks for the hint. – Bombyx mori Sep 1 '12 at 8:37
yeah ... kinda something like that. But instead of squaring ... you can take Real part of $z^2$ which makes the problem very simple. Kinda confused on that – hasExams Sep 1 '12 at 8:38
I am not sure how to deal with the bottom part, $(a+b\cos[\theta])$ is not easily treatable. – Bombyx mori Sep 1 '12 at 8:45

We should have


The first part of the definite integral is easily evaluated. The second part is the same as evaluating

$$\int \frac{\cos\theta}{a+b\cos\theta}d\theta=\int\frac{1}{b}\left(1-\frac{a}{a+b\cos\theta}\right)d\theta.$$

So we only need to evaluate


This can be done by various trignometry identities such as using $\tan(\theta/2)$. A detailed step by step proof can be found here (click show steps).

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thanks for idea ... I think i can evaluate $\frac{1}{a+b\cos[\theta]}d\theta$ very easily – hasExams Sep 1 '12 at 8:25
well, I am not sure if this is the "fastest possible" way to do it. I guess you are expecting some contour integral or manipulations involving $\Gamma$ functions. – Bombyx mori Sep 1 '12 at 8:28
I'm supposed to use contour integral ... the $z^2$ part is awful – hasExams Sep 1 '12 at 8:40

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\, \dd\theta:\ {\large ?}.\qquad\qquad a > b > 0}$.

\begin{align}&\color{#c00000}{% \int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta} =2\int_{0}^{\pi}{\cos^{2}\pars{\theta} \over a - b\cos\pars{\theta}}\, \dd\theta =-\,{2 \over b} \int_{0}^{\pi}{\cos^{2}\pars{\theta} \over \cos\pars{\theta} - \mu}\,\dd\theta \\[3mm]&\mbox{where}\quad\mu \equiv {a \over b}\,,\qquad \mu > 1 \end{align}

\begin{align}& \color{#c00000}{\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta} =-\,{2 \over b}\int_{0}^{\pi}{% \bracks{\cos\pars{\theta} - \mu}\bracks{\cos\pars{\theta} + \mu} + \mu^{2} \over \cos\pars{\theta} - \mu}\,\dd\theta \\[3mm]&=-\,{2 \over b}\int_{0}^{\pi}\bracks{\cos\pars{\theta} + \mu}\,\dd\theta -{2 \over b}\,\mu^{2}\int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - \mu} \end{align}

\begin{align}& \color{#c00000}{\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta} =-\,{2a \over b^{2}}\,\pi - {2a^{2} \over b^{3}} \color{#00f}{\int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - \mu}}\tag{1} \end{align}

With the usual Weierstrass Tangent Half-Angle substitution: \begin{align}&\color{#00f}{\int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - \mu}} =\int_{0}^{\infty}{2\,\dd t/\pars{1 + t^{2}}\over \pars{1 - t^{2}}/\pars{1 + t^{2}} - \mu} =-2\int_{0}^{\infty}{\dd t \over \pars{\mu + 1}t^{2} + \mu - 1} \\[3mm]&=-2\,{1 \over \mu - 1}\,\root{\mu - 1 \over \mu + 1}\ \overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}^{\ds{=\ {\pi \over 2}}} =-\,{\pi \over \root{\mu^{2} - 1}}\ =\ =-\,{b \over \root{a^{2} - b^{2}}}\,\pi \end{align}

Replacing in $\pars{1}$: $$ \color{#66f}{\large\int_{0}^{2\pi}{\cos^{2}\pars{\theta} \over a + b\cos\pars{\theta}}\,\dd\theta ={2a \over b^{2}}\pars{{a \over \root{a^{2} - b^{2}}} - 1}\pi}\,,\qquad a > b > 0 $$

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