It is said that the following integral equals 0 from some ADC circuit design book. I tried to prove it but did not work out. $$ \int_{0}^{\pi} \ln\left( \left| 1 + a e^{-jx} \right| \right) \, dx = 0 \quad \text{where} \quad |a| \leq 1 $$ Here, $a$ is real, and $j$ is the imaginary unit.
Can someone prove it or give some clues? Thanks a lot.
In the domain $\mathbb C\setminus(-\infty, 0]$, the principal value of $\ln$ defines a holomorphic function, i.e. $\ln (re^{j\theta})=\ln r + j\theta$ where $\theta\in(-\pi, \pi)$. Moreover, this satifies $\ln z_1z_2=\ln z_1 + \ln z_2$ as long as $$\operatorname{Arg} z_1,\operatorname{Arg} z_2\in (-\frac{\pi}{2}, \frac{\pi}{2})\Rightarrow \operatorname{Arg}(z_1z_2)\in(-\pi, \pi)$$ where $\operatorname{Arg}$ also takes the principal value.
Therefore we have $$\begin{align}\int_0^{\pi}\ln |1+ae^{-jx}|dx&=\int_0^\pi\frac{\ln(1+ae^{-jx})(1+ae^{jx})}{2}dx \\ &=\frac{1}{2}\int_0^\pi [\ln (1+ae^{-jx})+\ln (1+ae^{jx})]dx \\ &= \frac{1}{2}(I_-+I_+)\end{align}$$
So it suffices to show $I_++I_-=0$, where $I_{\pm}:=\int_0^{\pi} \ln (1+ae^{\pm jx})dx$.
We shall turn them into complex integrals in order to apply complex analysis.
When $x$ goes from $0$ to $\pi$, $z=e^{-jx}$ passes through the lower arc of the unit circle clockwise. And $dz= -jzdx$, $dx = -j\frac{dz}{z}$. Thus
$$I_- = j\int_{C_-} \frac{\ln(1+az)}{z}dz$$
where $C_-$ is the lower half of the unit circle oriented counterclockwise. The extra minus sign is because of the orientation.
Similarly, when $x$ goes from $0$ to $\pi$, $z=e^{jx}$ passes through the upper half of the unit circle counterclockwise (denoted as $C_+$), and $dx=j\frac{dz}{z}$, thus
$$I_+ = j\int_{C_+} \frac{\ln(1+az)}{z}dz$$
Now we have $$I_++I_- = j\int_{C}\frac{\ln (1+az)}{z}dz=0$$ because $\frac{\ln(1+az)}{z}$ has only a removable singularity at $z=0$ inside $C$.
(To make this more explicit we can use the power series expansion: $\ln (1+u) = u - \frac{u^2}{2}+\frac{u^3}{3}-\cdots$ for $|u|<1$. We can also apply the series expansion to $\ln (1+ae^{\pm jx})$ to compute $I_+, I_-$ directly and show they cancel each other.)
