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I need to prove that

\begin{equation} I_n=\displaystyle\int_{-1}^{1} \ln T_n^2(x) \frac{dx}{\sqrt{1-x^2}} = -2\pi\ln2 \end{equation} where $\ln$ denotes the natural logarithm and $T_n(x)$ are the Chebyshev polynomials of the first kind defined by $T_n(x) = cos(n\hspace{0.1cm}cos^{-1}(x))$. Performing the change of variable $x = \cos\theta$ one can easily obtain

\begin{equation} I_n=\displaystyle\int_{0}^{\pi} \ln \left\|\cos(n\theta)\right\|^2 d\theta \end{equation} using that $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ I can write

\begin{equation} I_n=\displaystyle\int_{0}^{\pi} \ln \cos^2(n\theta) d\theta = \frac{1}{2}\displaystyle\int_{0}^{2\pi} \ln \left\|\frac{e^{in\theta}+e^{-in\theta}}{2}\right\|^2 d\theta = \frac{1}{2}\displaystyle\int_{0}^{2\pi} \ln \left\|\frac{e^{2in\theta}+1}{2}\right\|^2 d\theta \end{equation} Now I use the mean value property for holomorphic functions which establishes that if $f$ is analytic in a region $D$ and $a\in D$, then $f(a)$ equals the integral around any circle centered at $a$ divided by $2π$ \begin{equation} f(a) = \frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}f(a+re^{i\theta})d\theta \end{equation}

In my case $f(a+re^{i\theta}) = \ln \left\|\frac{e^{2in\theta}+1}{2}\right\|^2$ so $a=r=1/2$. Then, the integral is

\begin{equation} \displaystyle\int_{0}^{2\pi} \ln \left\|\frac{e^{2in\theta}+1}{2}\right\|^2 d\theta = 2\pi f(a) = 2\pi \ln\left\|\frac{1}{2}\right|^2 = -2\pi\ln{4} = -4\pi \ln 2 \end{equation} and then \begin{equation} I_n=\displaystyle\int_{0}^{\pi} \ln \left\|\cos(n\theta)\right\|^2 d\theta = \frac{1}{2} \displaystyle\int_{0}^{2\pi} \ln \left\|\frac{e^{2in\theta}+1}{2}\right\|^2 d\theta = -2\pi\ln2 \end{equation} which is the desired result. I'm not sure if my procedure is correct because I'm working on a circle of radius $1/2$ centered at $1/2$ so my frontier contains the point $0$ for what the logarithm has an essential singularity. I'm not sure how to deal with that aspect. To what extent do I need to worry about what happens at the frontier?

solyav14
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    You should look at this answer I wrote on a related integral a while back. Firstly, you should notice that the function $f(x)$ isn't analytic (so you can't use mean value), but you should be able to manipulate it (as in my answer) to get an analytic function + some stuff that cancels when integrating. You'll still end up with a problematic singularity, but (as in my answer), the singularity here can be avoided by taking a limit as the radius $r\rightarrow 1/2^−$, though you should think about why such a limit is justified. – Christian E. Ramirez Nov 06 '22 at 04:16
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    If you're just interested in the integral and not in your particular proof method, you may also notice that by the periodicity and symmetry of the integrand, your second integral is just equal to $$4\int_0^{\pi/2}\log(\cos(x))dx$$This integral is very commonly asked about, and has very elementary solutions, for example here. – Christian E. Ramirez Nov 06 '22 at 04:26

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