7
$\begingroup$

How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?

My attempt:

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

To find the singularity, I solve: $ (2+\cos\theta)^2 = 0 $ and therefore, $\cos\theta = -2$.

Substituting: $\cos z = \frac{e^{iz} + e^{-iz}}{2} = \frac{z + \frac{1}{z}}{2}$, I find that $z = -2 + \sqrt{3} $ is the singular point that lies in the unit circle $|z| = 1$.

From this point, I have little idea how to go about solving this problem. I know I have to find the residue and then just sum them but to get the expression that would cancel out the pole is where I am currently stuck.

$\endgroup$
2
  • 1
    $\begingroup$ If you want to do it by residues, you will have to write it as a path integral with $dz$. $\endgroup$
    – GEdgar
    Feb 20 '13 at 2:29
  • $\begingroup$ Even if $|z| \leq 1$; it doesn't mean you have $\theta \in [-\pi, \pi]$ $\endgroup$
    – hjpotter92
    Feb 20 '13 at 2:40
8
$\begingroup$

I will start from the point where you left off. The integral may be written as, upon substitution of $z=e^{i \theta}$:

$$\begin{align}\frac{1}{2} \int_0^{2 \pi} \frac{d\theta}{(2+\cos{\theta})^2} &= -i 2 \oint_{|z|=1} dz \frac{z}{(z^2+4 z+1)^2}\\ &= -i 2 \oint_{|z|=1} dz \frac{z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2}\end{align}$$

after a bit of algebra. This integral is $i 2 \pi$ times the sum of the residues of the poles within the integration contour. The only pole inside this contour, as you point out, is the pole at $z=-2+\sqrt{3}$. The other pole at $z=-2-\sqrt{3}$ is outside this contour and is not counted.

To compute the residue at this pole, note that we have double roots. For such roots, we have to take a derivative:

$$\mathrm{Res}_{z=-2+\sqrt{3}} \frac{-i 2 z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2} = \lim_{z \rightarrow -2+\sqrt{3}} \left [\frac{d}{dz} \frac{-i 2 z}{(z+2+\sqrt{3})^2} \right ]$$

I will leave the algebra to the reader; the result is $-i \sqrt{3}/9$. The integral is then

$$\int_0^{\pi} \frac{d\theta}{(2+\cos{\theta})^2} = i 2 \pi \frac{-i \sqrt{3}}{9} = \frac{2 \sqrt{3}}{9} \pi$$

$\endgroup$
5
$\begingroup$

Trigonometric substitution:

$$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1}dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}\Longrightarrow$$

$$\int\limits_0^\pi\frac{d\theta}{(2+\cos\theta)^2}=\int\limits_0^\infty\frac{2\,dx}{1+x^2}\frac{1}{\left(2+\frac{1-x^2}{1+x^2}\right)^2}=2\int\limits_0^\infty\frac{x^2+1}{(x^2+3)^2}dx=$$

$$2\int\limits_0^\infty\left(\frac{1}{3+x^2}-\frac{2}{(3+x^2)^2}\right)=\left.\frac{2}{\sqrt 3}\arctan\frac{x}{\sqrt 3}\right|_0^\infty-\left.4\left(\frac{x}{6(3+x^2)}+\frac{1}{6\sqrt 3}\arctan\frac{x}{\sqrt 3}\right)\right|_0^\infty=$$

$$=\frac{4}{3\sqrt 3}\frac{\pi}{2}=\frac{2\pi}{3\sqrt 3}$$

$\endgroup$
3
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\pi}{\dd\theta \over \bracks{2 + \cos\pars{\theta}}^{\,2}}} = \left.-\,\totald{}{\mu}\int_{0}^{\pi}{\dd\theta \over \mu + \cos\pars{\theta}} \right\vert_{\ \mu\ =\ 2} \\[5mm] = &\ -\,\totald{}{\mu}\braces{\int_{0}^{\pi/2}{\dd\theta \over \mu + \cos\pars{\theta}} + \int_{-\pi/2}^{0}{\dd\theta \over \mu - \cos\pars{\theta}}}_{\mu\ =\ 2} \\[5mm] = &\ \left.-\,\totald{}{\mu}\int_{0}^{\pi/2}{2\mu \over \mu^{2} - \cos^{2}\pars{\theta}}\,\dd\theta\,\right\vert_{\ \mu\ =\ 2} \\[5mm] = &\ \left.-2\,\totald{}{\mu}\int_{0}^{\pi/2}{\mu \over \mu^{2}\sec^{2}\pars{\theta} - 1} \,\sec^{2}\pars{\theta}\,\dd\theta\,\right\vert_{\ \mu\ =\ 2} \\[5mm] = &\ \left.-2\,\totald{}{\mu}\int_{0}^{\infty}{\mu \over \mu^{2}t^{2} + \mu^{2} - 1} \,\dd t\,\right\vert_{\ \mu\ =\ 2} \\[5mm] = & \left.-\pi\,\totald{}{\mu}{1 \over \root{\mu^{2} - 1}}\right\vert_{\ \mu\ =\ 2} = \bbx{\ds{{2\root{3} \over 9}\,\pi}} \end{align}

$\endgroup$
2
  • $\begingroup$ this answer is really good, the only reason why it gets only two upvotes is being late in the party $\endgroup$
    – Shing
    May 15 '19 at 13:11
  • 1
    $\begingroup$ @Shing That's true. Sometimes old questions reappear in the timeline due to a MSE-Algorithm ( I guess so ). Thanks. $\endgroup$ May 15 '19 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.