# Showing $\int_0^\pi\ln\left|1+ae^{-jx}\right|\,dx=0$, where $|a|\leq1$, and $j$ is the imaginary unit

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.

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• What if $j=0$ and $q=1$? Commented Jun 11 at 8:26
• @Basics I suppose that, in this context, $j$ means $i$. Commented Jun 11 at 8:28
• Is $a\in\Bbb R$? Commented Jun 11 at 8:28
• a is real, j is the constant sqrt(-1) Commented Jun 11 at 8:32
• The answers to this question should be applicable here. Commented Jun 11 at 8:35

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 $$\int_0^{\pi}\ln |1+ae^{-jx}|dx=\int_0^\pi\frac{\ln(1+ae^{-jx})(1+ae^{jx})}{2}dx=\frac{1}{2}\int_0^\pi \ln [(1+ae^{-jx})+\ln (1+ae^{jx})]dx$$

So it suffices to show $$I:=I_1+I_2=0$$, where $$I_1==\int_0^{\pi} \ln (1+ae^{-jx})dx , I_2=\int_0^{\pi}\ln (1+ae^{jx}) dx$$.

We shall turn them into complex integrals to apply complex analysis.

When $$x$$ goest 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

$$I_1 = -j\int_{C_2} \frac{\ln(1+az)}{z}dz$$

where $$C_2$$ 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 $$C_1$$), and $$dx=-j\frac{dz}{z}$$, thus

$$I_2 = -j\int_{C_1} \frac{\ln(1+az)}{z}dz$$

Now we have $$I_1+I_2 = -j\int_{C}\frac{\ln (1+az)}{z}dz=0$$ because $$\frac{\ln(1+az)}{z}$$ has only a removable singularity at $$z=0$$.

(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 $$I_1, I_2$$ directly and show they cancel each other.)