Integrate: $\int_0^{\pi} \log ( 1 - 2 r \cos \theta + r^2)d\theta$

Question

If $r \in \Bbb R$ how to integrate $\displaystyle \int_0^{\pi} \log ( 1 - 2 r \cos \theta + r^2)d\theta$?

I need some hints. Special case, if $r = 1$ then I know the above integral is zero.

Here is my working \begin{align*} \int_0^{\pi}\log (1 - 2 r \cos \theta + r^2)d\theta &= \int_0^\pi\log ((1 - re^{i \theta})(1 -re^{-i\theta} )) d\theta\\ &= \int_0^\pi \log(1 - r e^{i\theta})d\theta + \int_0^\pi\log(1 - re^{-i\theta})d\theta\\ &= \int_0^{2\pi} \log( 1 - re^{i\theta})d\theta \\ &= 0 \end{align*}

Does this answer your question? Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) , dx$ — YuiTo Cheng, Aug 22 '20 at 14:40

score 6 · Accepted Answer · answered May 20 '13 at 15:58

6

Two hints:

The integrand is an even function of $\theta$ $\Rightarrow$ the integral can be written as $\frac12\int_0^{2\pi}$.
$1-2r\cos\theta+r^2=(1-re^{i\theta})(1-re^{-i\theta})$.

answered May 20 '13 at 15:58

Start wearing purple

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yeah I got that ... i got $0$, but if $|r| > 1$ , answer says that $2 \pi \log r$. Don't know if the answer is correct or not. – Mula Ko Saag May 20 '13 at 16:01
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then 3rd hint: $1-2r\cos\theta+r^2$ is also equal to $r^2\left(1-r^{-1}e^{i\theta}\right)\left(1-r^{-1}e^{-i\theta}\right)$. – Start wearing purple May 20 '13 at 16:03
this works ... but could you explain why two different cases? – Mula Ko Saag May 20 '13 at 16:06
because to expand the function $\ln(1-z)$ into Taylor series around $z=0$, you need $|z|<1$. Then you basically use that for any $n\neq0$ we have $\int_0^{2\pi}e^{in\theta}d\theta=0$. – Start wearing purple May 20 '13 at 16:08

score 2 · Answer 2 · answered May 20 '13 at 15:59

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Hint: Note that when $r<1$ this is the real part of $$\frac12 \int_{|z|=r} \log(1-z)^2\,\frac{dz}{iz}$$ for the usual branch of $\log$. When $r>1$, you're going to need to make a branch cut.

answered May 20 '13 at 15:59

Ted Shifrin

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hmmm ... how would i do it? $\log (1-z) = \ln |1 - z| + i \arg ( 1-z)$?? and why do I need to make cut if $r>1$ ? – Mula Ko Saag May 20 '13 at 16:07
No, you should use the Cauchy Integral Formula (or, if you prefer to think of it that way, the Residue Theorem). – Ted Shifrin May 20 '13 at 16:39

Integrate: $\int_0^{\pi} \log ( 1 - 2 r \cos \theta + r^2)d\theta$

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