# Poisson kernel for upper half plane

Can anyone tell me how to calculate the Poisson kernel for the upper half plane? I am able to calculate it for the unit disc and I know the unit disc and the upper half plane are conformally equivalent, do I need this?

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You probably don't need this, but I'd probably try to use the Cayley-transform to guess the Poisson kernel for the upper half plane. If you get the formula right, then it should be possible to derive it without resorting to the unit disk, but I don't quite see the point (and I haven't checked). Added: Actually Wikipedia seems to answer this question. Could you please try and be a bit more precise what exactly you're asking? I must say that I'm a bit confused. – t.b. May 12 '11 at 12:06

This is a good exercise. Let $\phi$ be the conformal mapping of the half plane to the unit disk.
To create a harmonic function on $\mathbb{H}$ which agrees with $f$ on the real line, one good strategy would be to translate it to the unit disk. Using the Poisson kernel for the disk, we can find a harmonic function on the disk which agrees with $f\circ \phi^{-1}$ on the boundary. Compose it with $\phi$ (which is also harmonic) to get a function which is harmonic on $\mathbb{H}$ that agrees with $f$ on the real line.
Given that you know how to calculate the Poisson kernel $p(z,t), z\in \mathbb{H}, t\in \mathbb{R}$for the upper half plane, the harmonic measure of an interval $I\subset \mathbb{R}$ w.r.t $z\in \mathbb{H}$ is given by the intergral of $p(z,t)$ over $I$. So one thing: you are calculating harmonic measure of a measurable subset of the boundary, NOT in the interior. – Mathmath Oct 6 '12 at 23:13