Prove that Poisson kernel integrates to $1$: $\frac{1}{2 \pi}\int_{-\pi}^{\pi} P(r,\theta - \Phi) d \Phi=1$

Question

I've been reading a lot about Poisson Kernel and there's always a property that I understand but can't proove.

First, $$P(r,\theta)=\frac{1-r^2}{1-2r\cos\theta+r^2}$$ and

$$U(r,\theta)=\frac{1}{2 \pi}\int_{-\pi}^{\pi} P(r,\theta - \Phi) f(\Phi) d \Phi$$

So, if we take $f(\Phi) =1$, I need to prove that $U(r, \theta)=1$, i.e.

\begin{align} U(r,\theta) = \frac1{2\pi} \int_{-\pi}^{\pi} \frac{(1-r^2)d\Phi}{1-2r \cos(\theta-\Phi)+r^2}=1 \end{align}

I was thinking that the proof could use the fact that $\Delta$U (r, $\theta$)=0, where $\Delta$ is the laplacian operator, but I don't know what to do with that. Can you help me? Thank you so much.

Answer 1

If $f(\Phi)=1$, the kernel integral is

\begin{align} U(r,\theta) = \frac1{2\pi} \int_{-\pi}^{\pi} \frac{(1-r^2)d\Phi}{1-2r \cos(\theta-\Phi)+r^2} = \frac1\pi \int_{0}^{\pi} \frac{(1-r^2)d\Phi}{1-2r \cos\Phi+r^2} \end{align}

Use $\cos\Phi=\frac2{\sec^2\frac{\Phi}2}-1 $

\begin{align} U(r,\theta) = & \frac1\pi \int_{0}^{\pi} \frac{(1-r^2) \sec^2\frac{\Phi}2}{(1+r)^2\sec^2\frac{\Phi}2-4r }d\Phi \\ = & \frac2\pi \int_{0}^{\pi} \frac{(1-r^2)d( \tan\frac{\Phi}2 )}{(1-r)^2+(1+r)^2\tan^2\frac{\Phi}2}\\ = & \frac2\pi \tan\left( \frac{1+r}{1-r}\tan\frac\Phi2 \right) \bigg|_0^\pi= 1 \end{align}