Is $\int_0^{\pi}\log \sin \theta d\theta$ not well-defined?

Question

Is $\int_0^{\pi}\log \sin \theta d\theta$ not well-defined?

On the third edition of Ahlfors' Complex Analysis, page 160 it states: As a final example we compute the special integral \begin{equation*} \int_0^{\pi}\log \sin \theta d\theta. \end{equation*} ...

The same proof applies near the vertex $\pi$, and we obtain \begin{equation*} \int_0^{\pi}\log (-2ie^{ix}\sin x)dx=0. \end{equation*}

Here are my questions.

(i) I think $\int_0^{\pi}\log \sin \theta d\theta$ is not well defined, since $\log \sin\theta$ is undefined when $\theta=0$ and $\theta=\pi$.

(ii) Actually, I think we only obtain something like \begin{equation*} \lim_{\delta\rightarrow 0^{+}}\int_{0+\delta}^{\pi-\delta}\log (-2ie^{ix}\sin x)dx=0. \end{equation*} Why is $\lim_{\delta\rightarrow 0^{+}}\int_{0+\delta}^{\pi-\delta}\log (-2ie^{ix}\sin x)dx=\int_0^{\pi}\log (-2ie^{ix}\sin x)dx=0$? Notice $\log (-2ie^{ix}\sin x)$ is undefined when $x=0$ and $x=\pi$.

I show you the whole context of the questions below. I have trouble understanding the statement with the red line.

Answer 1

As Lebesgue integral it's always well-defined because log is Lebesgue integrable on (0,1).

What's not defined is the log function on the complex plane. More precisely, the imaginary part of log will change 2pi if you draw a loop containing the origin. What the book is doing in some sense is to avoid this issue when taking contour integral.