# Integration problem: $\int_{-\pi}^{\pi} | \log( | 1 + \exp(- I \nu ) | ) | \mathrm{d}\nu < \infty$

Hello, I'm trying to bound an integral. I have a function $A(\nu) = | 1 + \exp(-I \nu) |$ (with $I$ being the imaginary unit) and I want to show that the condition (Paley-Wiener criterion for causality) applies

$$\int_{-\infty}^{\infty} \frac{|\log(A(\omega))|}{1+\omega^2} \mathrm{d}\omega < \infty$$

(log is the natural logarithm) I used a transformation from $\omega$ to $\nu$: $\omega = \tan(\nu/2)$ and I converted the integral by substitution to

$$\int_{-\pi}^{\pi} | \log(A(\nu)) |\mathrm{d}\nu < \infty$$

But I don't know how to show that this condition applies for the given function $A(\nu)$. I tried to simplify the problem by using $A(\nu) = | 1 + \exp(-I \nu) | = \sqrt{(1+\exp(-I\nu))(1+\exp(I\nu))}$ and thus simplifying the integral to

$$\frac{1}{2} \int_{-\pi}^{\pi} | \log(1+\exp(-I\nu)) + \log(1+\exp(I\nu)) |\mathrm{d}\nu$$

But still I have trouble finding a bound. I also tried $A(\nu) = | 1 + \exp(-I \nu) | = \sqrt{2} \sqrt{\cos(\nu) + 1}$.

I though maybe the problem can be solved by providing an upper and lower bound function that converges. But because $A(\nu)$ has values in the range $[0,1]$ the logarithm assumes very large values (and there are actually points of singularity for $A(\nu)=0$).

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You want to show that $$\int_{-\pi}^\pi|\log|1+e^{-it}||dt$$ is finite. Now $$|1+e^{-it}|=|e^{it/2}+e^{-it/2}|=2\cos(t/2)$$ so your integral is $$\int_{-\pi}^\pi|\log|2\cos(t/2)||dt =2\int_0^\pi|\log|2\cos(t/2)||dt.$$ Replacing $t$ by $\pi-2$ in the last integral gives $$2\int_0^\pi|\log|2\sin(t/2)||dt.$$ The integrand is nicely continuous away from $0$. Near $0$, $\sin (t/2)=tf(t)$ where $f(t)\to1/2$ as $t\to0$. Then the integrand is $|\log t+g(t)|$ where $g$ is continuous at $0$ and now finiteness follows since $$\int_0^1|\log t|dt$$ is finite (integration by parts).
Thank you very much for your answer. (I think you got a small typo. Replacing $t$ by $\pi-t$ and not $\pi-2$) But I have troubles following the last step of your argument. Why can I ignore the $g(t)$ for the proof of finiteness? –  user10256 Oct 22 '10 at 16:02
oh I think I understand it now. $log t$ is not continuous at $0$. But $g(t)$ is. So I don't have to worry about $g(t)$. It has no singularities and therefore is finite. The question left is whether the singularity of $log t$ causes the integral to be infinite. But it is easy to show that $\int_0^1 | \log t | \mathrm{d}t$ is finite and therefore the initial integral. Thanks thanks thanks. –  user10256 Oct 23 '10 at 21:52
The key is to understand the behavior of $A(\nu)$ near the singularity $\nu=0$. Using Taylor expansion we know that for $\nu$ small $A(\nu) = 1+e^{-I \nu} \approx -I\nu$. Therefore, $\log|A(\nu)| \approx \log|-I \nu| = \log |\nu|$. Note that $\int_{-\pi}^\pi \log|\nu| d\nu = 2 \int_{0}^\pi \log \nu d\nu < \infty$. To make this precise you need to control the error terms in the Taylor approximation.
There is no singularity at $\nu=0$, but at $\nu=\pm\pi$. For $nu$ small, $A(\nu)\approx2$. –  Julián Aguirre Oct 22 '10 at 14:58
Oops, as Julian points out I made a mistake above. To handle the singularities at $\pi$ and $-\pi$ you can still use a Taylor approximation. Near $\pi$ we have $A(\nu) \approx -I(\nu-\pi)$. Therefore the integral should be finite if $\int_0^\pi \log|\nu-\pi|\, d\nu < \infty$. –  Jon Peterson Oct 25 '10 at 14:15