# Integral of the conjugate Poisson Kernel

I'm dealing with a huge contradiction regarding Poisson's Kernel and conjugate Poisson's Kernel.

The situation is the following. Write the Poisson Kernel as $$P(z) = \frac{1-r^2}{1 + r^2 - 2r \cos \theta}, \ z = re^{i\theta}$$ and the conjugate Poisson Kernel as $$Q(z) = \frac{2r \sin \theta}{1 + r^2 - 2r \cos \theta}, \ z = re^{i\theta}$$ We know this functions are harmonic in the unit disk $\mathbb{D}$, and that the function $f:\mathbb{D} \longrightarrow \mathbb{C}$ defined as $f(z) = P(z) + i Q(z)$ is holomorphic.

Fixing $r < 1$, and applying Cauchy's integral theorem we get $$\int_{C(0,r)} f(z) dz = 0$$ Meaning that $$\int_{C(0,r)} P(z) dz = - i \int_{C(0,r)}Q(z) dz$$ Because $P(z)$ and $Q(z)$ are harmonic real valued functions, we have $$\int_{C(0,r)} P(z) dz = 2\pi r P(0) = 2\pi r$$ $$\int_{C(0,r)} Q(z) dz = 2\pi r Q(0) = 0$$ so $2\pi r = 0$.

What am I doing wrong? Sure it is something really basic, but I'm not seeing it.

Thank you very much!

## 1 Answer

We have $$\frac{1+z}{1-z} = \frac{1+re^{i\theta}}{1-re^{i\theta}} = \frac{1-r^2+2ir\sin{\theta}}{1+r^2-2r\cos{\theta}},$$ so the function is indeed analytic inside $D(0,1)$.

Your problem is that that the formula $$\int_{C(0,r)} P(z) \, dz = 2\pi r P(0)$$ is not true: starting from $$\frac{1}{2\pi i}\int_{C(0,r)} \frac{f(z)}{z} \, dz = f(0),$$ which is Cauchy's integral formula (note the difference being the division by $z$), and rewriting it using $dz = ire^{i\theta} d\theta$ gives Gauss's mean value theorem $$\frac{1}{2\pi r} \int_0^{2\pi} f(re^{i\theta}) \, d\theta = f(0),$$ and splitting this into real and imaginary parts gives $$\frac{1}{2\pi r} \int_0^{2\pi} P(re^{i\theta}) \, d\theta = P(0)$$ and the same with $P \to Q$. This is not the same as $\int_{C(0,r)} P(z) \, dz$, because working backwards, $d\theta=-i dz/z$.

• Omg, you're right. Thank you very much! Oct 19, 2017 at 11:22