How to evaluate $\ \int_{0}^{\pi} \log (|1 - a e^{-i w}|)\,dw $

Question

I massaged it into (assuming $ a = 1 $):

$$\ \int_{0}^{\pi} \log \left(2\left|\sin\left(\frac{w}{2}\right)\right|\right)\,dw $$

and I've seen how to solve for $\ \int_{0}^{\pi} \log(\sin (x)) \,dx$ over $[0 ,\pi]$. But I was unable to solve for this. I know the answer is 0 for $\ a\le1 $, and I have some intuition for why this is true (by graphing $\ |1 - a e^{-i w}| $ over $[0 ,\ \pi ]$ for different values of a), but I am not sure how to solve this analytically. This is from an engineering textbook on delta sigma data converters (eq. 4.26, pg. 99, Understanding Delta-Sigma Data Converters, 2nd Edition).

Answer 1

Assume that $a\in\mathbb{R}$. Then, observing that $|1-ae^{-iw}|=\sqrt{1+a^2-2a\cos(w)}$ is $2\pi$-periodic and even in $w$, we find that

$$\begin{align} \int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw&=\frac12 \int_0^{2\pi}\log\left(\sqrt{1+a^2-2a\cos(w)}\right)\,dw\\\\ &=\frac12 \int_0^{2\pi}\log\left(\sqrt{1+a^2+2a\cos(w)}\right)\,dw\\\\ &\overbrace{=}^{z=e^{iw}}\frac12 \oint_{|z|=1} \frac{\log\left(|z-a|\right)}{iz}\,dz\tag1 \end{align}$$

In addition, we see that

$$\int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw=\pi \log(|a|)+\frac12 \oint_{|z|=1} \frac{\log\left(|z-1/a|\right)}{iz}\,dz\tag2$$

It suffices, therefore, to assume that $a>1$.

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Now, for $a>1$, let us evaluate the integral $I(a)$ given by

$$I(a)=\frac12\oint_{|z|=1}\frac{\log(z-a)}{iz}\,dz$$

Inasmuch as $\log(z-a)$ is analytic in and on $|z|=1$, the Residue Theorem guarantees that

$$I(a)=\pi \log(a)+i\pi^2\tag3$$

where we have chosen to cut the plane with a ray from $a$ and extending along the positive real axis such that $\log(1)=0$.

We also can write

$$\begin{align} I(a)&=\frac12 \oint_{|z|=1}\frac{\log\left(|z-a|\right)}{iz}\,dz+\frac12 \oint_{|z|=1}\frac{\arg\left(z-a\right)}{z}\,dz\\\\ &= \int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw+\frac i2 \int_{-\pi}^\pi \left(\pi+\arctan\left(\frac{\sin(\phi)}{\cos(\phi)-a}\right)\right)\,d\phi\\\\ &= \int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw+i\pi^2\tag4 \end{align}$$

whence comparing $(3)$ and $(4)$ reveals

$$\int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw=\pi\log(a)$$

for $a>1$.

Using $(1)$, $(2)$ and $(4)$ we see that for $0<a<1$

$$\int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw=0$$

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Putting everything together yields

$$\int_0^\pi \log\left(|1-ae^{-iw}|\right)\,dw=\begin{cases}\pi\log(|a|)&,|a|\ge 1\\\\ 0&,|a|\le 1\end{cases}$$

Answer 2

Let $-1 < a < 1, \,z = e^{i w}$. Since $|1 - a \overline z| = |1 - a z|$, $$2 I(a) = \int_0^{2 \pi} \ln |1 - a e^{i w}| \,dw = \operatorname{Re} \int_0^{2 \pi} \ln(1 - a e^{i w}) \,dw = \operatorname{Re} \int_\gamma \frac {\ln(1 - a z)} {i z} dz = 0,$$ as there is a disk on which the integrand is analytic and which contains the unit circle $\gamma$.

Now let $a < -1 \lor a > 1$. Then $$I(a) = \int_0^\pi \ln {\left| a e^{-i w} \left( \frac {e^{i w}} a - 1 \right) \right|} \,dw = \pi \ln |a| + I {\left( \frac 1 a \right)} = \pi \ln |a|.$$ $I(\pm 1)$ can be found from $$I(a^2) = \frac 1 2 \int_0^{2 \pi} \ln |(1 - a e^{i w/2}) (1 + a e^{i w/2})| \,dw = 2 I(a).$$