A problem about the improper integral in Jensen's formula

Question

In Jensen's formula, when dealing with zeros of $f$ in the boundary of $D(0;r)$, we need the following lemma: $$ \int_{0}^{2\pi}\log\left(\left\vert\,1 - {\rm e}^{{\rm i}t}\,\right\vert\right)\,{\rm d}t = 0. $$

My question is how much one could understand this improper integral (e.g. Is it Lebesgue integrable?) As far as I know, apply Cauchy's theorem, we have its Cauchy principal value: $$ \lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{2\pi - \epsilon}\log\left(\left\vert\,1 - {\rm e}^{{\rm i}t}\,\right\vert\right)\,{\rm d}t = 0. $$

Answer 1

The singularity is of the form $\int_{-1}^1 \log|x| \, dx$. This is a very mild singularity in that $\int_{\epsilon}^1 |\log|x|| \, dx = \epsilon \log \epsilon - \epsilon + 1 $ converges as $\epsilon \to 0^+$. In particular, it is definitely Lebesgue integrable.

Answer 2

Since $1-e^{it}=(1-\cos t)-i\sin t$, $\sin^2t+\cos^2t=1$ and $1-\cos t=2\sin^2 \dfrac t2$ you want to prove that $$\int_0^{2\pi}\log \sqrt{2(1-\cos t)}dt=\int_0^{2\pi}\log {2\sin\frac t 2}dt=0$$

that is $$\frac 1 2\int_0^{2\pi}\log\sin\frac t 2dt=\int_0^\pi\log\sin tdt=-\pi\log 2$$

This is actually a known integral, and has been answered a few times here in MSE.

ADD The function is Lebesgue integrable, since $−\log\sin t$ is positive over [0,π/2] and bounded above by $−\log(2t/π)$ which is cleary Lebesgue integrable over $[0,π/2]$. You can use monotone convergence, and symmetry about $\pi/2$.