I am trying to evaluate the integral

$$\int_0^{2\pi} \frac{d\theta}{\left(1+\beta \cos \left(\theta\right)\right)^2}$$

via change of variables and applying Cauchy's Residue Theorem. Here is how I'm approaching this.

First, I make the the change of variables:

$$\begin{align*} z&=e^{i\theta}\\ dz&=\dfrac{1}{iz}d\theta\\ \cos\left(\theta\right)&=\dfrac{1}{2}(z+z^{-1}) \end{align*} $$

Substituting these back in, the integral then becomes the complex line-integral,


The integrand has three singularities: a removable singularity $z=0$ (that approaches 0), and poles of order 2 at $z=\dfrac{-1}{\beta}\left(1\pm\sqrt{1-\beta^2}\right)$. This means that if we just compute the sum of the residues of the integrand and multiply that by $2\pi i$, then by Cauchy's Theorem we'll have evaluated the primary integral.

To compute the residue of the integrand at $z=\dfrac{-1}{\beta}\left(1\pm\sqrt{1-\beta^2}\right)$, because there are poles of order 2 at each of the points we must first take the derivative of the part that is analytic in a nearby domain of the singularity, and then evaluate it at that point.

At this point it becomes extremely messy. I will edit this and write in my computations, but only if nobody seems to find any error or shortcut in my previous steps.

EDIT: As suggested in the comments, the substitution $z=\beta e^{i \theta}$ is also valid. Doing that, the integral becomes

$$\begin{align*} \oint_{|z|=\beta} \frac{dz}{iz\left(1+\frac{1}{2}\left(z+z^{-1}\right)\right)^2} &= -i\oint_{|z|=\beta}\frac{2z}{\left(z^2+2z+1\right)^2}dz\\ &=-i\oint_{|z|=\beta} \frac{2z}{\left(z+1\right)^4}dz \end{align*}$$

But look at that! The integrand is analytic in the region bounded by $|z|=\beta$, and so by Cauchy's Theorem the integral is always $0$, which can't be right. Any more suggestions?

  • $\begingroup$ Please do not use displaystyle maths (\dfrac triggers this) in question titles, so they don't take away unnecessary space from the front page. $\endgroup$
    – AlexR
    Nov 26, 2014 at 17:44
  • $\begingroup$ What about the substitution $z = \beta e^{i \theta}$? $\endgroup$
    – Crostul
    Nov 26, 2014 at 17:56
  • $\begingroup$ @AlexR I will remember that. $\endgroup$ Nov 26, 2014 at 18:39
  • $\begingroup$ @Crostul See the edit. $\endgroup$ Nov 26, 2014 at 18:41

1 Answer 1


One pole is inside the contour the other is outside the contour. And if you put $z = \beta\exp(i\theta)$ then $z^{-1}$ will have a factor of $\beta^{-1}$ instead of the desired $\beta$ so you must multiply that by $\beta^{2}$.

Note that the sum of all the residues in the complex plane, including the residue at infinity, always equals zero. This means that the value of a contour integral is given by either the sum of the residues of the poles inside the contour, or by minus the sum of the residues of the poles outside the contour.

  • $\begingroup$ I don't really understand what you said. Could you put in the context of what I wrote a bit more? (Sorry for my feeble mindedness) $\endgroup$ Nov 26, 2014 at 20:00
  • $\begingroup$ In your case the sum of the two residues equals zero, but only one of the poles is inside the contour, so you need to consider only the residue of that pole. This will be the one with the minus sign in your formula for the poles (because you must have that $|\beta|<1$). The simplest way to calculate the residue is to expand the function around that point. $\endgroup$ Nov 26, 2014 at 20:16
  • $\begingroup$ Okay I understand what you said, but there are two things I'd like to say. (1) "... the sum of all the residues in the complex plane, including the residue at infinity, always equals zero." I never knew that at all. You're sure that is true? (2) What went wrong in my approach using the substitution $z=\beta e^{i\theta}$ ? Changing the approach should still lead me to the correct answer, shouldn't it? $\endgroup$ Nov 27, 2014 at 18:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .