My book gives the Poisson Formula for such a harmonic function as:
$$ u(x + iy) = \frac{1}{\pi} \int_{-\infty}^{\infty}{\frac{y \cdot u(t) dt}{(t – x)^2 + y^2}} $$
Here is what I have attempted. First, I assume $ f $ is an analytic function s.t. $ u(x, y) = \text{Re}(f(x, y)) $. Then, I used this contour to integrate over:
So given any z, these is a big enough R s.t. the following is true by the Cauchy Integral Formula:
$$ f(z) = \frac{1}{2 \pi i} \int_{C}{\frac{f(\zeta)}{\zeta – z}d\zeta} $$
Also, $ z \in \text{upper half-plane} \implies \bar{z} \in \text {lower half-plane} $, which tells us that:
$$ 0 = \frac{1}{2 \pi i} \int_{C}{\frac{f(\zeta)}{\zeta – \bar{z}}d\zeta} $$
Subtracting the second from the first yeilds:
$$ f(z) – 0 = f(z) = \frac{1}{2 \pi i} \int_{C}{\frac{f(\zeta)}{\zeta – z} – \frac{f(\zeta)}{\zeta – \bar{z}}d\zeta} = \frac{1}{2 \pi i} \int_{-R}^{R}{\frac{2iy \cdot f(\zeta)}{(\zeta – z)(\zeta – \bar{z})}d\zeta} + \int_{C_R}{\frac{2iy \cdot f(\zeta)}{(\zeta – z)(\zeta – \bar{z})}d\zeta} $$
This is where I am stuck. I cannot figure out why:
$$ \int_{C_R}{\frac{2iy \cdot f(\zeta)}{(\zeta – z)(\zeta – \bar{z})}d\zeta} = 0 $$
Thanks for the help.
Best Answer
If you assume that $f = u+iv$ is bounded, say $|f| \le M$, you can argue like this: Since $|(\zeta - z)(\zeta-\bar z)| > \frac12R^2$ on $C_R$ for $R$ large, we get
$$ \left|\int_{C_R}{\frac{2iy \cdot f(\zeta)}{(\zeta - z)(\zeta - \bar{z})}d\zeta} \right| \le \frac{4yM}{R^2} \cdot \pi R \to 0 $$ as $R\to\infty$ by the "standard estimation lemma" for complex integrals.
It is possible to reduce the problem to this case by approximating $u$ with functions that are harmonic on a neighbourhood of the closed upper half-plane and then the corresponding $f$'s are automatically bounded on the upper half plane. You may find it easier to map the upper half-plane conformally to the unit disc and use that version of Poisson's integral formula and then map back.
I would be happier with assuming that $u$ extends continuously to the real axes. Otherwise you have to worry about what you actually mean by the integral.