Using the method of residues, verify the following.
$$\int_0^{\pi} \frac{d \theta}{(3+2cos \theta)^2} = \frac{3 \pi \sqrt{5}}{25}$$

I tried doing this but can't get the correct answer, here is my attempt:
first I substituted $cos \theta = \frac{z+ \frac{1}{z}}{2}$ and $d \theta = \frac{dz}{iz}$ and got $\int_0^{\pi} \frac{dz}{iz(3+z+\frac{1}{z})^2}$

then I multiplied top and bottom my $iz$ so that the $i$ moves to the top and on the bottom I'll have $z^2$ so I can distribute it into the rest of it and get $-i \int_0^{\pi} \frac{zdz}{(3z+z^2+1)^2}$

the denominator has two zeroes at $-\frac{3}{2} \pm \sqrt{\frac{5}{4}}$, since the denominator is squared the function has poles here of order two.

The I used Cauchy's Residue theorem but first I needed to find the residue at $-\frac{3}{2} + \sqrt{\frac{5}{4}}$ because only this pole is inside the unit circle.

I found the residue using theorem 1 on pdf page 324/579 from this textbook http://english-c.tongji.edu.cn/_SiteConf/files/2014/05/05/file_53676237d7159.pdf

After finding the residue using that formula and applying cauchy's residue theorem I do not get an answer close to $\frac{3 \pi \sqrt{5}}{25}$

I am not sure how to handle the fact that the integral is from $0$ to $\pi$ rather then from $0$ to $2\pi$, and I am not sure if I made any mistakes when finding the poles and residues.

  • 2
    $\begingroup$ Application of the residue theorem requires a closed contour. Starting with the original integral, note that the integrand is even. Thus, $\int_0^\pi =\frac12 \int_{-\pi}^\pi$. $\endgroup$ – Mark Viola May 16 '17 at 16:22
  • $\begingroup$ @MarkViola, ok, does it look like I found the correct poles and that they are both of order 2 because that's where I might have made a mistake, that was a typo I had $(3z+z^2+1)^2$ on my paper when doing it $\endgroup$ – idknuttin May 16 '17 at 16:24
  • $\begingroup$ Yes, the poles are correct. $\endgroup$ – Mark Viola May 16 '17 at 16:26
  • 1
    $\begingroup$ Well I did it your way (after using the periodicity) and crunched out the residue and got the answer stated. $\endgroup$ – ancientmathematician May 16 '17 at 16:31
  • $\begingroup$ @MarkViola You should turn your comment into an answer. $\endgroup$ – zhw. May 16 '17 at 16:46

Note $$\int_0^{\pi} \frac{d \theta}{(3+2cos \theta)^2} = \frac12\int_0^{2\pi} \frac{d \theta}{(3+2cos \theta)^2}.$$ Let $z=e^{i\theta}$ and hence one has \begin{eqnarray} &&\int_0^{\pi} \frac{d \theta}{(3+2cos \theta)^2}\\ &=&\frac12\int_0^{2\pi} \frac{d \theta}{(3+2cos \theta)^2}\\ &=&\frac12\int_{|z|=1}\frac{1}{(3+z+z^{-1})^2}\frac{dz}{iz}\\ &=&\frac12\int_{|z|=1}\frac{z}{(z^2+3z+1)^2}dz\\ &=&\pi\text{Res}(f(z),z=\frac{1}{2}(-3+\sqrt5))\\ &=& \frac{3 \pi \sqrt{5}}{25} \end{eqnarray} where $f(z)=\frac{z}{(z^2+3z+1)^2}$ has a pole at $z=\frac{1}{2}(-3+\sqrt5)$.

  • $\begingroup$ thank you, I kept checking over my work and realized I made an algebraic mistake. $\endgroup$ – idknuttin May 16 '17 at 17:08
  • $\begingroup$ You're welcome. $\endgroup$ – xpaul May 16 '17 at 17:26

By the substitution $\theta=2\varphi$ and the cosine duplication formula we have

$$ I = \int_{0}^{\pi}\frac{d\theta}{(3+2\cos\theta)^2} = 2\int_{0}^{\pi/2}\frac{d\varphi}{(1+4\cos^2\varphi)^2} \tag{1}$$ and by the substitution $\varphi=\arctan t$ the problem boils down to evaluating $$ 2\int_{0}^{+\infty}\frac{(1+t^2)}{(5+t^2)^2}\,dt = \int_{-\infty}^{+\infty}\frac{(1+t^2)}{(5+t^2)^2}\,dt.\tag{2}$$ The meromorphic function $f(t)=\frac{(1+t^2)}{(5+t^2)^2}$ has a double pole at $t=\pm i\sqrt{5}$ and behaves like $\frac{1}{t^2}$ for $|t|\to +\infty$. By the residue theorem it follows that:

$$ I = 2\pi i\,\text{Res}(f(t),t=i\sqrt{5}) =2\pi i\lim_{t\to i\sqrt{5}}\frac{d}{dt}\frac{(1+t^2)}{(t+i\sqrt{5})^2}=\color{red}{\frac{3\pi}{5\sqrt{5}}}\tag{3}$$ as wanted, after a straightforward computation.

There also is a simple geometric approach. $\rho(\theta)=\frac{p}{1+\varepsilon\cos\theta}$ is the polar equation of an ellipse with respect to a focus. Since the area in polar coordinates is given by $\frac{1}{2}\int_{0}^{2\pi}\rho(\theta)^2\,d\theta$, $I$ just depends on the area of an ellipse ($\pi ab $) with a given eccentricity and a given semi-latus rectum. This proves the more general $$ \int_{0}^{\pi}\frac{d\theta}{(u+v\cos\theta)^2} = \frac{\pi u}{\left(u^2-v^2\right)^{3/2}} \tag{4}$$ as soon as $0<v<u$.


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.