Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \begin{align} \int_0^{2\pi}\log|e^{i\theta} - 1|d\theta &= \int_0^{2\pi}\log(1-\cos(\theta))d\theta \\ &= \int_0^{2\pi}\log(\cos(0) - \cos(\theta))\,d\theta\\ &= \int_0^{2\pi}\log\left(-2\sin\left(\frac{\theta}{2}\right)\sin\left(\frac{-\theta}{2}\right)\right)\,d\theta\\ &= \int_0^{2\pi}\log\left(2\sin^2\left(\frac{\theta}{2}\right)\right)\,d\theta\\ &= \int_0^{2\pi}\log(2)d\theta + 2\int_0^{2\pi}\log\left(\sin\left(\frac{\theta}{2}\right)\right)\,d\theta\\ &= 2\pi \log(2) + 4\int_0^\pi \log\big(\sin(t)\big)\,dt\\ &=2\pi \log(2) - 4\pi \log(2) = -2\pi \log(2) \end{align} $$

Where $\int_0^\pi \log(\sin(t))\,dt = -\pi \log(2)$ according to this. The first step where I removed the absolute value signs is the one that worries me the most. Thanks.

share|cite|improve this question
up vote 3 down vote accepted

Don't forget De Moivre's formula!

$$\begin{array}{c l} |e^{i\theta}-1| & =|(\cos\theta-1)+i\sin\theta| \\[2pt] & =\sqrt{(\cos\theta-1)^2+\sin^2\theta} \\ & = \sqrt{(\cos^2\theta+\sin^2\theta)+1-2\cos\theta} \\ & = \sqrt{2-2\cos\theta}.\end{array}$$

Don't worry though,

$$\log \sqrt{2-2\cos\theta}=\frac{\log2+\log(1-\cos\theta)}{2}$$

so there's not too much you need to modify in your computation.

share|cite|improve this answer
Ok cool, thanks anon. – Thoth Apr 21 '12 at 18:37
Well, not much to modify in the computation, but it changes the result rather significantly, namely to zero :-) – joriki Feb 28 '13 at 10:56

You can use Jensen's Formula, I believe:

Edit: Jensen's formula seems to imply that your integral is zero...

share|cite|improve this answer
log|z| has a singularity at zero, thus it's not holomorphic in a neighborhood containing our curve. – Thoth Apr 21 '12 at 20:17
The conditions in Jensen's Theorem apply to $f$, not to $\log f$. If you integrate around a circle of radius 1/2, Jensen's Theorem is definitely applicable. – Jim Apr 22 '12 at 1:23
Applying the correction in anon's answer also leads to zero. – joriki Feb 28 '13 at 10:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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