How to prove $\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0$? How can I prove $\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0$, where $a<1$?
Thanks.
 A: Denote by $z=ae^{ix}$ then $dz=aie^{ix}dx\Rightarrow dx=\frac{dz}{iz}$. Your integral can be rewritten in the following form:
$$\int^{2\pi}_{0}\ln(1+a^2+2a\cos(x))\,dx=\oint_{|z|=a}\ln[(z+1)(\bar{z}+1)]\frac{dz}{iz}$$
The integrand has a simple pole at $z=0$ with residue $0$. Therefore
$$\oint_{|z|=a}\ln[(z+1)(\bar{z}+1)]\frac{dz}{iz}=2\pi i\cdot \text{Res}\left(\frac{\ln[(z+1)(\bar{z}+1)]}{iz},0\right)=2\pi i\cdot 0=0 $$
A: if the integral is $f(a)$ then $f(0)=0$. 
also
$$
1+a^2+2a \cos x = (1+ae^{ix})(1+ae^{-ix})
$$
so 
$$
f(a) = \int_0^{2\pi} \ln(1+ae^{ix}) dx +  \int_0^{2\pi} \ln(1+ae^{-ix}) dx
$$
$$
\frac{df}{da} = \int_0^{2\pi} \frac{e^{ix}}{1+ae^{ix}} dx + \int_0^{2\pi} \frac{e^{-ix}}{1+ae^{-ix}} dx \\
=0
$$
since both integrands are analytic for $a \lt 1$
A: Let $a$ be a real number.


*

*$\color{blue}{\text{Case 1.}}$  $\quad|a|<1$



$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=0
$$ 

Observe that 
$$
\left(1+ae^{ix}\right)\left(1+ae^{-ix}\right)=1+2a\cos x+ a^2, \quad x \in [0,2\pi],
$$
and that
$$
\begin{align}
\log \left(1+2a\cos x+ a^2\right)
&=\log \left(1+ae^{ix}\right)+\log \left(1+ae^{-ix}\right)\\\\
&=-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n e^{inx}-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n e^{-inx}\\\\
&=-2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \cos (nx)\\\\
\end{align}
$$
Using the normal convergence of the series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \cos (nx) $ as a function of $x \in [0,2\pi]$,
$$
\left|\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \cos (nx)\right|\leq \sum_{n=1}^{\infty}\frac{|a|^n }{n}=-\log(1- |a|)<\infty,
$$
we are allowed to perform a termwise integration giving 
$$
\begin{align}
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right) {\rm d}x
=-2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \int_0^{2\pi}\cos (nx) dx=0
\end{align}$$
due to
$$
\begin{align}
\int_0^{2\pi}\cos (nx) {\rm d}x=\left. \frac{\sin (nx)}{n}\right|_0^{2\pi}=0, \quad n=1,2,3,\ldots.
\end{align}$$


*

*$\color{blue}{\text{Case 2.}}$  $\quad|a|>1$



$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=4\pi \log |a|
$$ 

Observe that 
$$
\log \left(1+2a\cos x+ a^2\right)=2\log |a| +\log \left(1+2\cdot\frac1a\cos x+ \frac{1}{a^2}\right), \quad x \in [0,2\pi],
$$
then
$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=2\log |a|\int_0^{2\pi} {\rm d}x+\int_0^{2\pi}\log \left(1+2\cdot\frac1a\cos x+ \frac{1}{a^2}\right){\rm d}x, 
$$ and, applying the previous case to the last integral, we get the desired result.
Remark 1. 
Here $\displaystyle  \log (z)$ denotes the principal value of the logarithm defined for $z \neq 0$ by 
$$ \begin{align} 
  \displaystyle \log (z)  = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi. 
\end{align}   
$$
Remark 2. 
$$
\begin{align}
& a=-1 & \text{gives} \quad &\int_0^{2\pi}\log \left(2-2\cos x\right){\rm d}x=4\int_0^{\pi}\log \left(2\sin u\right){\rm d}u=0\\
& a=1 & \text{gives} \quad &\int_0^{2\pi}\log \left(2+2\cos x\right){\rm d}x=2\int_0^{\pi}\log (4\cos^2 u){\rm d}u=0.
\end{align}
$$
Remark 3. One may readily notice that the preceding reasoning gives

$$
\begin{align} \int_0^{\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x =
\left\{
 \begin{array}{ll}
  0  & \mbox{if } |a| \leq 1 \\
  2\pi \log |a| & \mbox{if } |a| > 1
 \end{array}
\right.
\end{align}   
$$

