Maximum entropy spectral analysis

APPENDIX B

We need to compute the integral

$$\frac{1}{2\pi}\int_{-\pi}^\pi \ln\left[\exp(i\theta) - Z_0\right]... ... \frac{P.V.}{2\pi i}\oint_{\vert Z\vert = 1} \ln\left(Z-Z_0\right)\frac{dZ}{Z},$$ (B-1)

where $P.V.$ stands for the principal value of the contour (complex) integral when the logarithm's branch cut is taken along the negative real axis.

First, note that

$$\frac{1}{2\pi}\int_{-\pi}^\pi \ln\left[\exp(i\theta) - Z_0\right]... ...+ \frac{1}{2\pi}\int_{-\pi}^\pi \ln\left[1 - \exp(-i\theta)Z_0\right]\,d\theta.$$ (B-2)

The first integral on the right is just

$$-\frac{1}{2\pi i}\int_{-\pi}^\pi \theta d\theta = 0,$$ (B-3)

since the integrand is an odd function. When $\vert Z_0\vert < 1$, the integrand of the second integral on the right can be expanded in a convergent power series. Integrating term by term, we find that

$$-\frac{1}{2\pi}\int_{-\pi}^{\pi} \sum_{n = 1}^\infty\frac{Z_0^n}{n}\exp\left(i n\theta\right)d\theta = 0,$$ (B-4)

since $\exp\left(in\pi\right) - \exp\left(-in\pi\right) = \cos(n\pi) +i\sin(n\pi) - \cos(-n\pi) -i\sin(-n\pi) = 0$ (the two cosines cancel and the two sines both vanish individually for all integer values of $n$). Thus, we find (B-1) is identically zero for all $\vert Z_0\vert < 1$. In particular, it vanishes when $Z_0 = 0$, so

$$\frac{P.V.}{2\pi i}\oint_{\vert Z\vert = 1} \ln Z \times \frac{dZ}{Z} = 0.$$ (B-5)

Maximum entropy spectral analysis