How to prove that $\int_{[-\pi,\pi]}\log(\vert 1- \exp(it)\vert)\mathrm{d}\lambda(t)=0$?

Solution 1:

The following observation is useful:

$$ \log |1 - re^{it}| \geq \log |1 - e^{it}| \qquad\text{for} \quad t \in \mathbb{R} \text{ and }r > 1 $$

Indeed,

$$ \frac{\partial}{\partial r} \log |1 - re^{it}| = \frac{\partial}{\partial r} \operatorname{Re}\log(r - e^{-it}) = \operatorname{Re}\left(\frac{1}{r - e^{-it}}\right) = \frac{r - \cos t}{|r - e^{-it}|^2} > 0, $$

and so, $r \mapsto \log |1 - re^{it}|$ is increasing for each given $t$.

Now, since $\log |1 - re^{it}|$ is squeezed between two integrable functions and converges pointwise to $\log|1 - e^{it}|$, we can apply the dominated convergence theorem to conclude.

Solution 2:

Let $r_n \to 1$

You can assume that $r_n<2,\forall n \in \Bbb{N}$

$|1-r_ne^{it}|\leq 1+r_n<3$ almost everywhere

Thus $\log|...| \leq \log 3 \in L^1[-\pi,\pi]$