Show (without using 3.8.9 in the S&S) that \[ \int_0^{2\pi} \log\abs{1 - e^{i\theta}}~d\theta = 0 \]
Show that this identity is equivalent to S&S 3.8.9: \[ \int_0^1 \log(\sin(\pi x)) ~dx = -\log 2 .\]
Solution
For \(0<r<1\), the function \[ z\longmapsto \log(1-rz) \] is holomorphic on the unit disk, using the branch determined by its power series at \(0\). Its real part is \[ \log|1-rz|. \] By the mean-value property for harmonic functions, \[ \frac1{2\pi}\int_0^{2\pi} \log|1-re^{i\theta}|\,d\theta =\log|1|=0. \] Hence \[ \int_0^{2\pi} \log|1-re^{i\theta}|\,d\theta=0 \] for every \(0<r<1\).
Now let \(r\uparrow1\). Pointwise for \(0<\theta<2\pi\), \[ \log|1-re^{i\theta}| \longrightarrow \log|1-e^{i\theta}|. \] To justify passing to the limit, restrict to \(r\ge1/2\). Since \[ |1-re^{i\theta}|^2 =(1-r)^2+4r\sin^2(\theta/2), \] the negative part near the endpoints is bounded by a constant multiple of \[ 1+|\log\theta|+|\log(2\pi-\theta)|, \] which is integrable, while the positive part is uniformly bounded because \[ |1-re^{i\theta}|\le2. \] Dominated convergence therefore gives \[ \boxed{ \int_0^{2\pi} \log|1-e^{i\theta}|\,d\theta=0.} \]
For part (b), use \[ |1-e^{i\theta}| =2\sin(\theta/2), \qquad 0<\theta<2\pi. \] Thus part (a) is equivalent to \[ 0 =\int_0^{2\pi} \left(\log2+\log\sin(\theta/2)\right)d\theta. \] With the substitution \[ x=\frac{\theta}{2\pi}, \] this becomes \[ 0 =2\pi\log2 +2\pi\int_0^1\log(\sin\pi x)\,dx. \] Hence \[ \boxed{ \int_0^1\log(\sin\pi x)\,dx=-\log2.} \] Every step is reversible, so the two identities are equivalent.