TITLE: Trouble with Poisson integral
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: 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