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

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.

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2 Answers 2

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.

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Ok cool, thanks anon. –  Blood Pudding 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: http://mathworld.wolfram.com/JensensFormula.html

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

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log|z| has a singularity at zero, thus it's not holomorphic in a neighborhood containing our curve. –  Blood Pudding 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

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