Integrate $\int_{0}^{2\pi}\log|1-e^{i\theta}| d\theta$

Question

Here is exercise 11, chapter 3 from Stein & Shakarchi's Complex Analysis II:

Show that if $|a|<1$, then:

$$\int_{0}^{2\pi}\log|1-ae^{i\theta}|\,d\theta = 0$$

Then, prove that the above result remains true if we assume only that $|a|\leq 1.$ ($a\in \mathbb{C}$)

I've already proved it for $|a|<1$, but couldn't make it for $|a|=1$, which I thought would be the easy part. But the only thing I could come up with was that, WLOG, it is enough to prove that:

$$\int_{0}^{2\pi}\log|1-e^{i\theta}|\,d\theta = 0$$

Tried to look for some kind of symmetry, but could't find it. Any ideas? Thanks!

Answer 1

I'll plod on and see what happens.

We want to show that $\int_{0}^{2\pi}\log|1-e^{it}|dt = 0 $.

$\begin{array}\\ |1-e^{it}| &=|1-\cos(t)-i\sin(t)|\\ &=\sqrt{(1-\cos(t))^2+\sin^2(t)}\\ &=\sqrt{1-2\cos(t)+\cos^2(t)+\sin^2(t)}\\ &=\sqrt{2-2\cos(t)}\\ &=\sqrt{2(1-\cos(t))}\\ &=\sqrt{2(2\sin^2(t/2))}\\ &=2\sin(t/2) \qquad\text{if }0 \le t \le 2\pi\\ \end{array} $

so $\log|1-e^{it}| =\log(2\sin(t/2)) $ so that

$\begin{array}\\ \int_{0}^{2\pi}\log|1-e^{it}|dt &=\int_{0}^{2\pi}\log(2\sin(t/2))dt\\ &=2\pi\log(2)+\int_{0}^{2\pi}\log(\sin(t/2))dt\\ &=2\pi\log(2)+2\int_{0}^{\pi}\log(\sin(t))dt\\ &=2\pi\log(2)+2(-\pi \log(2)) \qquad\text{according to Wolfy}\\ &=0\\ \end{array} $

By Googling for "integral with sine and log" I got a number of references to $\int_{0}^{\pi}\log(\sin(t))dt =-\pi \log(2) $ being a result of Euler including this one:

http://www.ams.org/journals/proc/2005-133-05/S0002-9939-04-07863-3/S0002-9939-04-07863-3.pdf

(Added later)

According to that paper, Euler proved his result by using $ \log(\sin x) = -\sum_{n=1}^{\infty} \dfrac{\cos(2nx)}{n}− \log 2 $ and $\int_0^{\pi} \cos(2nx)dx = 0 $.

Answer 2

Note that for $|a|=1$, we can write $a=e^{i\psi}$. Then, exploiting the $2\pi$-periodicity of the integrand, we have

$$\begin{align} \int_{-\pi}^\pi \log|1-ae^{i\theta}|\,d\theta&=\int_{-\pi}^\pi \log|1-e^{i(\theta+\psi)}|\,d\theta\\\\ &=\int_{-\pi+\psi}^{\pi+\psi} \log|1-e^{i\theta}|\,d\theta\\\\ &=\int_0^{2\pi}\log|1-e^{i\theta}|\,d\theta \end{align}$$

METHODOLOGY 1:

Note that we have $|1-e^{i\theta}|=\sqrt{2(1-\cos(\theta))}=2|\sin{\theta/2}|$. Now, we have

$$\int_0^{2\pi}\log|1-e^{i\theta}|\,d\theta=4\int_0^{\pi/2}\log(2\sin(\theta))\,d\theta=0$$

since $\int_0^{\pi/2}\log(\sin(\theta))\,d\theta=-\frac{\pi}{2}\log(2)$

---

METHODOLOGY 2:

Now, we cut the complex plane with a line from $(1,0)$ and extending along the positive real axis.

Note that $\log(1-z)$ is analytic within and on a closed contour $C$ defined by $z=e^{i\phi}$ for $\epsilon \le \phi \le 2\pi -\epsilon$, and $z=1+2\sin(\epsilon/2) e^{i\nu}$ for $\pi/2 + \gamma \le \nu \le 3\pi/2 -\gamma$, where $ \cos(\gamma)=\frac{\sin \epsilon}{\sqrt{2(1-\cos \epsilon)}}$ and $0 \le \gamma <2\pi$ on this branch of $\gamma$.

Then, from the residue theorem, we have

$$\int_C \frac{\log(1-z)}{z}dz=2\pi i \log(1-0)=0$$

which implies

$$\begin{align} \int_C \frac{\log(1-z)}{z} dz&=\int_{\epsilon}^{2\pi-\epsilon} \log(1-e^{i\phi})i d\phi+\int_{3\pi/2-\gamma}^{\pi/2+\gamma} \frac{\log(-2\sin(\epsilon/2) e^{i\nu})}{1+2\sin(\epsilon/2) e^{i\nu}}i2\sin(\epsilon/2) e^{i\nu}d\nu\\\\ &=i\int_{\epsilon}^{2\pi-\epsilon} \log(1-e^{i\phi}) d\phi+ i2\sin(\epsilon/2) \int_{3\pi/2-\gamma}^{\pi/2+\gamma} \frac{\log(-2\sin(\epsilon/2)e^{i\nu})e^{i\nu}d\nu}{1+2\sin(\epsilon/2)e^{i\nu}} \\\\ &=i\int_{\epsilon}^{2\pi-\epsilon} \log|1-e^{i\phi}| d\phi -i \int_{\epsilon}^{2\pi-\epsilon} \arctan \left(\frac{\sin \phi}{1-\cos \phi}\right)d\phi \\\\ &+ i2\sin(\epsilon/2) \int_{3\pi/2-\gamma}^{\pi/2+\gamma} \frac{\log(-2\sin(\epsilon/2)e^{i\nu})e^{i\nu}d\nu}{1+2\sin(\epsilon/2)e^{i\nu}} \\\\ &=0 \end{align}$$

As $\epsilon \to 0$ the first term on the RHS approaches $i$ times the integral of interest. The second term approaches zero since $\arctan(\frac{\sin \phi}{1-\cos \phi})$ is an odd, $2\pi$-periodic function of $\phi$ and the integration extends over the entire period. And the last term approaches $0$ since $x\log x \to 0$ as $x \to 0$. Thus, the integral of interest is zero!