Estimate the Bergman kernel on the unit disk Ok so I want to estimate the Bergman kernel and get something like
$\int_\mathbb{D} \frac{dV(\zeta)}{\lvert 1-z\bar{\zeta}\rvert ^2} \sim \frac{C \log (1-\lvert z \rvert ^2 )}{\lvert z \rvert ^2}$ on the unit disk.
So far, I have calculated the Bergman kernel on the unit disk itself and I know that the Bergman kernel on the unit disk is of the form $\frac{1}{\pi (1-z \bar{\zeta})^2}$.
On the other hand, there's a theorem apparently that states $\log (\frac{1}{K(z,z)}) \leq c - \log (1-\lvert z \rvert ^2)$, where $K(z,z)$ is the Bergman kernel on the unit disk. However, I cannot connect all this information and estimate the Bergman kernel the way I want.
So far, I think it is something that I should probably compute using the polar coordinates, but don't have any clue where would this take me to. Any help is appreciated.
 A: Denote the Bergman kernel by $K_z(\zeta)$. Notice that the integral you  want to calculate is simply the norm of the reproducing kernel (probably multiplied by some constant). The reproducing kernel property always implies $$ \|K_z(\zeta)\|^2 = \int_{\mathbb{D}}K_z(\zeta)\overline{K_z(\zeta)} \,dV(\zeta)=K_z(z)$$
hence the estimates of $K_z(z)$ are helpful.
A: First of all notice that $\int_{\mathbb{D}} |K_z(\zeta)|\, dV(\zeta) = \int_\mathbb{D} |K_{|z|}(\zeta)|\, dV(\zeta)$ due to the rotation invariance. Let us calculate $\int_\mathbb{D} |K_{r}(\zeta)|\, V(\zeta)$ for $r\in (0,1)$. Let the polar coordinates of $\zeta$ be $(\rho, \phi)$ then
$$
I_r = \int_\mathbb{D} \frac{1}{|1 - z\bar {\zeta}|^2}\, V(\zeta) = \int_{0}^1\int_0^{2\pi}\frac{\rho}{1 - 2r\rho\cos\phi + r^2\rho^2}\,d\phi\, d\rho. 
$$
Let us denote $r\rho$ by $a$ then we can calculate the inner integral in the following way:
$$
\int_0^{2\pi}\frac{d\phi}{1 - 2a\cos\phi + a^2} = \int_{-\pi}^{\pi}\frac{d\phi}{1 - 2a\cos\phi + a^2} = [u = \tan(\phi/2)]
\\
= \int_{-\infty}^{\infty}\frac{2/(u^2 + 1)}{1 - 2a(1 - u^2)/(1 + u^2) + a^2}du = \int_{-\infty}^{\infty}\frac{2}{u^2(a + 1)^2 + (1 - a)^2}du = \frac{2\pi}{1 - a^2}.
$$
And we see that
$$
I_r = 2\pi\int_0^1\frac{\rho\,d\rho}{1 - r^2\rho^2} = 2\pi\frac{-\log(1 - r^2)}{2r^2},
$$
which gives the required asymptotics.
