# How to prove the closed form of an integral

How to prove the following identity: $${1\over 2\pi}\int_{0}^{2\pi}\ln(1-2r\cos x+r^{2})\,dx=2\ln r,\text{ where } r\gt 1.$$

I was asked to use the following result to prove it: $$\lim_{n\to\infty}\sqrt[n]{f_{1n}f_{2n}\cdots f_{nn}}=\exp\left\{{1\over{b-a}}\int_a^b \ln f(x)\,dx\right\}$$ provided $f\in C[a,b]$, $f\gt 0$, and $f_{vn}=f(a+v\delta_n), \delta_n={{b-a}\over n}$

It's easy to prove the hint by checking the definition of Riemann integral,but I was trapped in using it to prove the original identity.

Thanks for help!

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Solution $1:$ note $(z-e^{i\alpha})(z-e^{-i\alpha})=z^2-2z\cos\alpha+1$ hence:

$$\big(z^{n}-1\big)^2=\prod_{k=1}^n \Big( z^2-2z\cos \left(\frac{2k}{n}\pi\right)+1\Big)\quad (\star)$$

Writing out the hint:

$$\lim_{n\to\infty} \sqrt[n]{\prod_{k=1}^n \Big( z^2-2z\cos \left(\frac{2k}{n}\pi\right)+1\Big)}=\exp\left\{\frac{1}{2\pi}\int_0^{2\pi}\ln \big(z^2-2z\cos t+1\big)\,\mathrm{d}t\right\}$$

Given $(\star)$ and $n\to\infty:\;\sqrt[n]{\big(z^{n}-1\big)^2}\to z^2$, the result follows.

Alternative approaches:

Solution $2:$ $\text{Log}\,z$ is holomorphic on $\mathbb{C}\setminus \mathbb{R}_{\leq 0}$, in particular on $\{z : |z −r| =1,\;r\in\mathbb{R}_{>1}\}$. Tweeking Gauss' MVT:

$$\ln r=\frac{1}{2\pi}\int_0^{2\pi}\text{Log} (r-e^{it})\,\mathrm{d}t\quad\text{and}\quad \ln r=\frac{1}{2\pi}\int_0^{2\pi}\text{Log} (r-e^{-it})\,\mathrm{d}t$$

Adding both yields the result.

Solution $3:$ as requested, below is a sketch for a solution which presumes no knowledge of $\mathbb{C}.$

$$\text{Define for}\;\lambda\geq 1:\;f(\lambda)=\int_0^{2\pi}\ln \big(\lambda^2-2\lambda\cos x+1\big)\,\mathrm{d}x\Rightarrow \frac{\mathrm{d}f}{\mathrm{d}\lambda}=\int_0^{2\pi}\cdots=\frac{4\pi}{\lambda}$$

The latter can be done using Weierstraß and is straightforward. Furthermore,

\begin{aligned}\frac{1}{2}f(1)-\pi\ln 2 &=\int_0^{\pi}\ln (1\pm\cos x)\mathrm{d}x\\&=\underbrace{\int_0^{\pi}\ln \sin x\,\mathrm{d}x}_{=\,-\pi\ln 2\;(\clubsuit)}\Rightarrow f(1)=0\end{aligned}\;\;\Longrightarrow\;\; \frac{1}{2\pi}f(r)=\int_1^{r}\frac{2\,\mathrm{d}\lambda}{\lambda}=2\ln r

$($the claimed result in $(\clubsuit)$ is a classic$)$

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Excellent answer!But can we prove it directly without using complex analysis? –  C Weid Dec 26 '13 at 17:17

Hint From your own definitions, (take the log of both sides of the hint and use $[a,b]=[0,2\pi]$): $$\begin{split} \frac{1}{2\pi} \int_0^{2\pi} \ln \left( 1-2r\cos(x)+r^2 \right) dx &= \lim_{n \to \infty} \ln \left( \sqrt[n]{\prod_{k=1}^n \left(1-2r\cos(2\pi k/n) + r^2 \right)} \right) \end{split}$$

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