Computing $\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy$

Question

If $\phi_\alpha(z)=(\alpha-z)/(1-\bar\alpha z)$ for $|\alpha|<1$, prove that

$\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy=1$ and $\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1-|\alpha|^2}{|\alpha|^2}log\frac{1}{1-|\alpha|^2}$ where in the case $\alpha=0$ the expression on the right is understood as the limit as $|\alpha|\to 0$.

[Hint: The first integral can be evaluated without a calculation. For the second, use polar coordinates, and for each fixed $r$ use contour integration to evaluate the integral in $\theta$]

I did the first question, but I have trouble doing the second question. I don't know how to use polar coordinates method here.

So I computed $\phi'_\alpha(z)=\frac{-1+\alpha\bar\alpha}{(1-\alpha z)^2}$ and by using polar coordinate method I know, I get $\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0|\frac{-1+\alpha\bar\alpha}{(1-\bar\alpha re^{i\theta)^2}}|rdrd\theta$, but it looks wrong. Can you give me any idea? Thank you.

Answer 1

$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0\frac{|-1+\alpha\bar\alpha|}{|1-\bar\alpha re^{i\theta)^2}|}rdrd\theta=(1-|\alpha|^2)\int^1_0r(\frac{1}{\pi}\int^{2\pi}_0\frac{1}{|1-\bar\alpha re^{i\theta)^2}|}d\theta)dr$

Clearly the integral is invariant under a circle rotation $\alpha \to \alpha e^{it_0}$ so we can assume $\alpha=c=|\alpha|>0$

$\frac{1}{|1-cre^{i\theta}|^2}=\frac{P(cr,\theta)}{1-c^2r^2}$ so $\frac{1}{\pi}\int^{2\pi}_0\frac{1}{|1-\bar\alpha re^{i\theta)^2}|}d\theta=\frac{2}{1-c^2r^2}=\frac{2}{1-|\alpha|^2r^2}$ as the integral of the Poisson kernel (divided by $2\pi$) is $1$ hence we get:

$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=2(1-|\alpha|^2)\int_0^1\frac{r}{1-|\alpha|^2r^2}dr=\frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$ so we are done!

Answer 2

You can directly compute the integral without using the Poisson kernel. We assume that $\alpha$ is real. Using $$ z=e^{i\theta}, \bar z=e^{-i\theta} $$ we have \begin{eqnarray} &&\frac{1}{\pi}\iint_\mathbb{D}|\phi'_\alpha|dxdy\\ &=&\frac{1-|\alpha|^2}{\pi}\int^1_0\int^{2\pi}_0\frac{1}{|1-\alpha re^{i\theta}|^2}r\,d\theta\,dr\tag1\\ &=&\frac{1-|\alpha|^2}{\pi}\int^1_0r\int_{|z|=1}\frac{1}{|1-\alpha rz|^2}\frac{dz}{iz}dr\tag2\\ &=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\int_{|z|=1}\frac{1}{(1-\alpha rz)(1-\alpha r\frac1z)}\frac{dz}{z}dr\tag3\\ &=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\int_{|z|=1}\frac{1}{(1-\alpha rz)(z-\alpha r)}dz\,dr\\ &=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\cdot2\pi i\frac{1}{1-\alpha^2r^2}dr\\ &=&2(1-|\alpha|^2)\pi\int^1_0\frac{r\,dr}{1-\alpha^2r^2}\\ &=&-\frac{1-|\alpha|^2}{\alpha^2}\ln(1-\alpha^2). \end{eqnarray} Update:

1. Since $z=e^{i\theta}$, we have $dz=ie^{i\theta}d\theta=izd\theta$ and hence $d\theta=\frac{dz}{iz}$. So we have (2).

2. Since $|z|^2=1$, we have $\bar z=\frac1z$ and hence $$|1-\alpha r z|^2=(1-\alpha r z)(1-\alpha r\bar z)=(1-\alpha z)(1-\alpha r\frac1z). $$ So we have (3).