Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$

Question

$\newcommand{\on}[1]{\operatorname{#1}}$ $$ \mbox{Show that for}\ r \in \left(-1,1\right),\ \int_{0}^{\pi}\log\left(1 - 2r\cos\left(t\right) + r^{2}\right){\rm d}t = 0 $$ Here's what I did so far: $$ \on{f}\left(r,t\right) = \log\left(1 - 2r\cos\left(t\right) + r^{2}\right) = \log\left(\left[1 - r{\rm e}^{{\rm i}t}\right] \left[1 - r{\rm e}^{-{\rm i}t}\right]\right) $$ The Leibniz Rule states that $$ \dfrac{\rm d}{{\rm d}r} \int_{0}^{\pi}\on{f}\left(r,t\right){\rm d}t = \int_{0}^{\pi} \dfrac{\partial\on{f}\left(r,t\right)} {\partial r}\,{\rm d}t $$ After calculating the right part I found $2\pi r$ which means that $$ \int_{0}^{\pi}\on{f}\left(r,t\right){\rm d}t = \pi r^2\ \mbox{when it should be}\ 0 $$ Thanks in advance.

Answer 1

Let $ n\in\mathbb{N} $, first of all, we wanna factor the polynomial $ X^{2n}-1 $, it has $ 2n-1 $ zeros which are $ \mathrm{e}^{\mathrm{i}\frac{k\pi}{n}} ,\ k\in\left[\!\left[0,2n-1\right]\!\right] $. Thus : \begin{aligned}X^{2n}-1=\prod_{k=0}^{2n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}&=\left(X-1\right)\prod_{k=1}^{n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\left(X+1\right)\prod_{k=n+1}^{2n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\\ &=\left(X^{2}-1\right)\prod_{k=1}^{n}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)}\prod_{k=1}^{n-1}{\left(X-\mathrm{e}^{\mathrm{i}\frac{\left(2n-k\right)\pi}{n}}\right)}\\ &=\left(X^{2}-1\right)\prod_{k=1}^{n}{\left(X-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}\right)\left(X-\mathrm{e}^{-\mathrm{i}\frac{k\pi}{n}}\right)}\\ X^{2n}-1&=\left(X^{2}-1\right)\prod_{k=1}^{n-1}{\left(X^{2}-2X\cos{\left(\frac{k\pi}{n}\right)}+1\right)}\end{aligned}

Hence, if $ r\in\left(-1,1\right) $, we have : $$ \prod_{k=1}^{n-1}{\left(r^{2}-2r\cos{\left(\frac{k\pi}{n}\right)}+1\right)}=\frac{r^{2n}-1}{r^{2}-1} $$

Using Riemann sum theorem, we have the following : \begin{aligned}\int_{0}^{\pi}{\ln{\left(r^{2}-2r\cos{x}+1\right)}\,\mathrm{d}x}&=\lim_{n\to +\infty}{\frac{\pi}{n}\sum_{k=0}^{n-1}{\ln{\left(r^{2}-2r\cos{\left(\frac{k\pi}{n}\right)}+1\right)}}}\\ &=\lim_{n\to +\infty}{\left(\frac{2\pi}{n}\ln{\left(1-r\right)}+\frac{\pi}{n}\ln{\left(\prod_{k=1}^{n-1}{\left(r^{2}-2r\cos{\left(\frac{k\pi}{n}\right)}+1\right)}\right)}\right)}\\ &=\lim_{n\to +\infty}{\left(\frac{2\pi}{n}\ln{\left(1-r\right)}+\frac{\pi}{n}\ln{\left(\frac{1-r^{2n}}{1-r^{2}}\right)}\right)}\\ \int_{0}^{\pi}{\ln{\left(r^{2}-2r\cos{x}+1\right)}\,\mathrm{d}x}&=0\end{aligned}

Answer 2

Note that $1 - 2r\cos t + r^2 = (1-re^{it})(1-re^{-it})$

\begin{align} \int_0^\pi &\log( 1 - 2r\cos t + r^2) dt =2\Re\int_0^\pi \log(1-re^{it}) dt\\ =& - 2\Re \int_0^\pi \sum_{k=1}^\infty \frac{(re^{it} )^k }kdt = -2\sum_{k=1}^\infty \frac{r^k}k \int_0^\pi \cos kt\ dt =0 \end{align}

Answer 3

If $I(r) $ is the integral then we have $I(0)=0$ and we can show that $I'(r) =0$ for all $r\in(-1,1)$ and then we get $I(r) =0$ for all $r\in(-1,1)$.

Clearly we have $$I'(r) =\int_{0}^{\pi}\frac{2r-2\cos t} {1-2r\cos t +r^2}\,dt$$ Clearly this is $0$ for $r=0$. So let $r\neq 0$ and then we can write $$I'(r) =\frac{1}{r}\int_0^{\pi}\left(1+\frac{r^2-1}{1-2r\cos t+r^2}\right)\,dt$$ and this equals $$\frac{1}{r}\left(\pi+\frac{\pi(r^2-1)}{\sqrt {(1+r^2)^2-4r^2}}\right)=0$$ as $r^2<1$.

We have used the standard formula $$\int_0^{\pi}\frac{dx}{a+b\cos x} =\frac{\pi} {\sqrt {a^2-b^2}},a>|b|$$

Answer 4

I used the naive method and from your insight $f(r,t)=\log(1-e^{it}r)+\log(1-e^{-it}r)$. Then we have, by substituting the inner integrand to be $u$ for each integral,

\begin{equation} \begin{split} \int_0^\pi\log(1-e^{it}r)dt+\int_0^\pi \log(1-e^{-it}r) dt&=\int_{1-r}^{1+r}\log u\dfrac{du}{i(u-1)}\\ &+\int_{1-r}^{1+r}\log u\left(-\dfrac{du}{i(u-1)}\right)=0 \end{split} \end{equation} as desired.

Answer 5

Doubling the integral, we may integrate over $[-\pi,\pi)$. The function $u(z)=\log(1-z)$ is analytic in the unit disk, so averaging it on the circle $\{|z|=r\}$ yields $u(0)=0$.

Answer 6

hint

$$1-2r\cos(t)+r^2=$$ $$(r-\cos(t))^2+\sin^2(t)$$

Answer 7

By substitution $(z=e^{it})$ and computing residues, we have the well known integrals:

$$J_1 = \int_0^{2\pi} \frac{dt}{1-2r\cos t+ r^2} = \frac{2\pi}{1-r^2}, \quad (|r|<1),$$ $$J_2 = \int_0^{2\pi} \frac{\cos t\, dt }{1-2r\cos t+ r^2} = \frac{2\pi r}{1-r^2}, \quad (|r|<1).$$

Let $$J_0(r) = \int_0^{2\pi} \log \left( {1-2r\cos t+ r^2} \right) dt$$

$$\frac{dJ_0}{dr} = \int_0^{2\pi} \frac{\partial \log \left( {1-2r\cos t+ r^2} \right)}{\partial r}dt=2 \int_0^{2\pi} \frac{r-\cos t}{ {1-2r\cos t+ r^2} }dt=rJ_1-J_2=0.$$ $$J_0(r) = \text{const}.$$ Since $J_0(0)=0,$ we have $$J_0(r) = 0,$$ but this is twice the integral we want, which is thus zero.

Answer 8

In my post, I found that $$ \int_{0}^{\pi} \ln (b \cos x+c) d x=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right) $$ where $\left|\frac{b}{c}\right| \leqslant 1$ and $c \neq 0$. $$ \begin{aligned} \int_{0}^{\pi} \ln \left(1-2 r \cos t+r^{2}\right) d t =& \pi \ln \left(\frac{1+r^{2}+\sqrt{\left(1+r^{2}\right)^{2}-(-2 r)^{2}}}{2}\right) = 0 . \end{aligned} $$