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Show that for $r \in (-1,1)$

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

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$ The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.

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    $\begingroup$ Already asked here (and there... and I've used it myself here). The whole bunch of beautiful answers below should go to the first linked question IMO, but... it's too late ;) $\endgroup$
    – metamorphy
    Jan 12, 2021 at 4:57
  • $\begingroup$ If you show your calculation of "the right part", then we can find your error for you. $\endgroup$
    – GEdgar
    Jul 3, 2022 at 16:48

8 Answers 8

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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}

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    $\begingroup$ +1 for evaluating by definition. Never expected complicated integrals to be handled this way. $\endgroup$
    – Paramanand Singh
    Jan 11, 2021 at 14:25
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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\\ = &\ 2Re \int_0^\pi \log(1-re^{it}) dt = - 2Re \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}

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  • $\begingroup$ Thanks a lot !! But how did you get from the 3rd to the 4th line? I see it's a geometric series but i can't figure it out exactly how. $\endgroup$
    – zartos
    Jan 11, 2021 at 8:12
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    $\begingroup$ @Zelda - it is the Taylor expansion of $\log(1-x)$ $\endgroup$
    – Quanto
    Jan 11, 2021 at 13:09
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    $\begingroup$ @Zelda Thanks to the $1/k$, it's not geometric, although it can be proved by integration of a geometric series. $\endgroup$
    – J.G.
    Jan 11, 2021 at 14:45
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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.

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  • $\begingroup$ Thanks! i never thought it could've been solved by a simple substitution! $\endgroup$
    – zartos
    Jan 11, 2021 at 8:25
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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 $u(0)=0$.

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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 $$\int_0^{\pi}\frac{dx}{a+b\cos x} =\frac{\pi} {\sqrt {a^2-b^2}},a>|b|$$

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hint

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

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  • $\begingroup$ I'm sorry...It is not possible $\ddot \smile$. $\endgroup$
    – Sebastiano
    Jan 10, 2021 at 23:58
  • $\begingroup$ Bettega is from the old Rubentus (another way of saying that Juventini steal championships) ahahahaah. $\endgroup$
    – Sebastiano
    Jan 11, 2021 at 0:07
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    $\begingroup$ Buenas noches italiano Sebastiano milano Torino Gullit Baggio. $\endgroup$ Jan 11, 2021 at 1:00
  • $\begingroup$ ahahhahahah :-) $\endgroup$
    – Sebastiano
    Jan 11, 2021 at 8:58
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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.

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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 $c \neq 0$. $$ \begin{aligned} \int_{0}^{\pi} \ln \left(1-2 r \cos t+r^{2}\right) d t =& \pi \ln \left(\frac{1+r^{2}+\sqrt{\left(1+r^{2}\right)^{2}-(-2 r)^{2}}}{2}\right) = 0 . \end{aligned} $$

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