I'm continuing my studies about the space $\mathbb{T}$ and I reach the point in which are introduced the Harmonic functions. Well up to now I have a little trouble with understanding the Poisson's integral. I know that this particular integral is defined as:
DEF: If $\mu\in\ M(\mathbb{T})$ we define the Poisson integral of $\mu$ by the expression for $z= re^{i\theta}$: $P(\mu)(z)=\int_{-\pi}^{\pi}P(ze^{-it})d\mu (t)$.
Now to understand better how to use it, I am trying doing some exercise about this Poisson integral. In particular, I'm trying to calculate $P(\mu)$ for this two cases:
For $\mu=\delta_{0}-\delta_{\pi/2}$ where $\delta_{x}$ is the Dirac measure for the points
$d\mu=D_{N}(t)dm(t)$ where $D_{N}$ is the kernel of Dirichelet.
I've tried to apply directly my definition but something goes wrong!
Can someone help me?
Note that the easiest way to do this type of question is to use either the harmonic representation of $P$ (for term by term integration):
$P(z)=\sum_{n <0}r^{|n|}e^{in\theta}+\sum_{n \ge 0}r^{n}e^{in\theta}=1+\sum_{n \ge 1}(\bar z^n+z^n), z=re^{i\theta}$
or the closed-form of $P$ (for substitution)
$P(z)=\Re \frac{1+z}{1-z}=\frac{1-r^2}{1-2r\cos \theta +r^2}$
Then by definition the first integral is $P(z)-P(ze^{-i\frac{\pi}{2}})=\frac{1-r^2}{1-2r\cos \theta +r^2}-\frac{1-r^2}{1-2r\sin \theta +r^2}$
while the second is:
$\frac{1}{2\pi}\int_{-\pi}^{\pi}(\sum_{n <0}r^{|n|}e^{in\theta}e^{-int}+\sum_{n \ge 0}r^{n}e^{in\theta}e^{-int})(\sum_{-N}^{N}e^{ikt})dt=$
$=\sum_{-N \le n <0}r^{|n|}e^{in\theta}+\sum_{0 \le n \le N}r^{n}e^{in\theta}=1+\sum_{1}^{N}(\bar z^n + z^n)$,
which is the $N$ truncation of it