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

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 $$0$$.

• 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 ;) Jan 12, 2021 at 4:57
• If you show your calculation of "the right part", then we can find your error for you. Jul 3, 2022 at 16:48

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}

• +1 for evaluating by definition. Never expected complicated integrals to be handled this way. Jan 11, 2021 at 14:25

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}

• 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. Jan 11, 2021 at 8:12
• @Zelda - it is the Taylor expansion of $\log(1-x)$ Jan 11, 2021 at 13:09
• @Zelda Thanks to the $1/k$, it's not geometric, although it can be proved by integration of a geometric series.
– J.G.
Jan 11, 2021 at 14:45

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,

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

• Thanks! i never thought it could've been solved by a simple substitution! Jan 11, 2021 at 8:25

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 $$u(0)=0$$.

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

hint

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

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

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.

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}