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 $A(\nu)=0$).

Please help me solve this problem.