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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  • 1
    $\begingroup$ What if $j=0$ and $q=1$? $\endgroup$
    – Basics
    Commented Jun 11 at 8:26
  • 3
    $\begingroup$ @Basics I suppose that, in this context, $j$ means $i$. $\endgroup$ Commented Jun 11 at 8:28
  • $\begingroup$ Is $a\in\Bbb R$? $\endgroup$ Commented Jun 11 at 8:28
  • $\begingroup$ a is real, j is the constant sqrt(-1) $\endgroup$
    – metroidman
    Commented Jun 11 at 8:32
  • 2
    $\begingroup$ The answers to this question should be applicable here. $\endgroup$
    – Matt L.
    Commented Jun 11 at 8:35

1 Answer 1


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.)


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