Integral of logarithm of Poisson kernel I would like to understand how should I compute using complex analysis the following integral,
$$I = \oint_{|z|=1}dz\,\ln\frac{1-|w|^2}{|w-z|^2},\quad |w|<1.$$
My attempt was following,
$$\ln\frac{1-|w|^2}{|w-z|^2}=\ln\left(1-|w|^2\right)-\ln\left|w-z\right|^2$$
and next notice that the second term is nothing more than also-called Poisson integral,
$$|w-z|^2=1-2r\cos(\theta-\phi)+r^2,\quad w=re^{i\phi},z=e^{i\theta}.$$
However, I assume that I am wrong because the first term is also gives zero. Could anyone please remind how to deal with following integral?
Addendum: Okey, it seems indeed zero. Let me check it in Wolfram Mathematica. For simplicity, let me use the following parametrization $w=re^{i\phi}$, $z=e^{i\theta}$, so
$$\ln\frac{1-|w|^2}{|w-z|^2}=\ln\frac{1-r^2}{1-2r\cos(\phi-\theta)+r^2}.$$
The corresponding integral becomes
$$\oint_{|z|=1}dz\ln\frac{1-|w|^2}{|w-z|^2}=\int_{-\pi}^{+\pi}d\theta\,e^{i\theta}i\ln\frac{1-r^2}{1-2r\cos(\phi-\theta)+r^2}\equiv J(r,\theta).$$
Let me evaluate the following cells,
J[r_, \[Phi]_] := 
 NIntegrate[
  Exp[I*\[Theta]]*I*
   Log[(1 - r^2)/(1 - 2*r*Cos[\[Phi] - \[Theta]] + 
       r^2)], {\[Theta], -Pi, +Pi}];

Jvalues = Table[{r*Cos[\[Phi]], r*Sin[\[Phi]], Abs[J[r, \[Phi]]]}, {r, 0, 
   0.95, 0.05}, {\[Phi], -Pi, +Pi, 2*Pi/20}];

ListPlot3D[Flatten[Jvalues, 1]]

I explicitly plot Abs because I feel that the imaginary part exists. The result of last cell is:

Unfortunately, Mathematica works extremely slow for such integrals, so I cannot use Integrate. Nevertheless, let me check a specific case, $\phi=0$:
Integrate[
 Log[(1 - r^2)/(1 - 2*r*Cos[\[Phi] - \[Theta]] + r^2)], {\[Theta], -Pi, +Pi},
Assumptions -> {0 < r < 1}]

The result for this integration coincides with my integral up to multiplication by $i$. Wolfram Mathematica says:
$$-\frac{2i\pi^2}{3}+4\pi\ln r + i \ln r^2 + 2\pi\ln(1-r^2)+2i\text{Li}_{2}(r)+2i\text{Re}\,\text{Li}_{2}(1/r).$$
As was noticed in comments,
$$\ln(1-2r\cos(\phi-\theta)+r^2)=-2\sum_{n=1}^{\infty}\frac{r^n}{n}\cos\left[n(\phi-\theta)\right].$$
It seems that the integral over $\theta$ for this expression is not zero,
$$\int_{-\pi}^{+\pi}d\theta\,e^{i\theta}\cos\left[n(\phi-\theta)\right]=\frac{2\sin(\pi n)e^{in\phi}}{n^2-1}.$$
Next, perform series summation,
$$-2i\sum_{n=1}^{\infty}\frac{r^n}{n}\frac{2\sin(\pi n)e^{in\phi}}{n^2-1}=2i\pi re^{i\phi}$$
So, it seems that integral is not zero.
 A: After an easy simplification the given integral is equal to  $I=∮_{|z|=1}(-2)ln|z-w|dz$.
Now, let $w=δe^{i\phi}$, $δ<1$, be the constant and $z=e^{iθ}$ be the paremetrization of the unit circle.
Then $ I=∫_0^{2π}(-i)ln[1+δ^2-2δcos(θ-φ)]e^{iθ}dθ$.
Here, we may use (5) of https://www.maa.org/sites/default/files/pdf/cms_upload/321916735744.pdf.bannered.pdf
We get $ I=2i∑_{n=1}^{∞}\frac{δ^n}{n}∫_0^{2π}cos(n(θ-φ))e^{iθ}dθ$,
And $I=i∑_{n=1}^{∞}\frac{δ^n}{n}∫_0^{2π}(e^{i(n+1)θ-ni\phi}+e^{i(1-n)θ+ni\phi})dθ$
But those integrals inside the summation are all zero excpet the integral $\int_0^{2\pi}e^{i(1-n)θ+ni\phi}d\theta$ when $n=1$, which evaluates to $2\pi e^{i\phi}$.
Hence, $I=2\pi i\delta e^{i\phi}=2\pi i w$.
Note: For the naive "complex" solution, I thought about the integral of logarithm function around the unit disk. I saw this post: What is the integral of log(z) over the unit circle?.
Now, I will try to make up a solution out of this mess. So far, I didn't like my solution at all.
The integral of this problem can be written in the form:
$$ I=-∮_{|z|=1}ln(z-w)dz-∮_{|z|=1}ln(\bar{z}-\bar{w})dz.$$
And further,
$$ I=-∮_{|z|=1}ln(z)dz-∮_{|z|=1}ln(\bar{z})dz-∮_{|z|=1}ln(1-\frac{w}{z})dz-∮_{|z|=1}ln(1-\frac{\bar{w}}{\bar{z}})dz.$$
Now, first two integrals cancel by our choice of fixed paremetrization. The third and the last integrals must be $2\pi i w$ and $0$ respectively.
