Poisson Integral Formula

June 2, 2014

This note contains a discussion of the Poisson Integral Formula. It will rely on some things we did in class this quarter and last quarter.

Let ϕ be Riemann integrable [0, 2π]. Define u(z) for |z| < 1 by

u(z) = \frac{1}{2π} Re (∫_{0}^{2π} \frac{(ζ + z)}{(ζ − z)}ϕ(t)dt) .

This expression can be written as a contour integral

u(z) = \frac{1}{2π} Re (∫_{0}^{2π} \frac{(ζ + z)}{(ζ − z)} \frac{ϕ(ζ)}{iζ}dζ) .

For most functions ϕ this integral cannot be evaluated explicitly. But, what if ϕ is identically 1? Then we

can use the residue theorem. The integrand has simple poles at z and 0. The residues are \frac{2}{i}, \frac{−1}{i}. So 2πi times the sum of the residues is 2π. So in this case

u(z) = 1,

as we found in class by other means.