Derive the Poisson Formula for a bounded C-harmonic function in the upper half-plane. 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.
 A: 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.
A: I would not want to introduce the (potentially unbounded) harmonic conjugate of $u$ into consideration. Here is another idea.
Let $$U(x + iy) = \frac{1}{\pi} \int_{-\infty}^{\infty}{\frac{y \cdot u(t) dt}{(t - x)^2 + y^2}} =\frac{1}{\pi} \int_{-\infty}^{\infty} u(t)\cdot \operatorname{Im} \frac{1}{t - x - iy} \,dt$$
For each fixed $t$ the function $x+iy\mapsto \operatorname{Im} \frac{1}{t - x - iy}$  is harmonic in the upper half-plane $H$. Therefore, so is $U$ (one way to prove this is to integrate $U$ over a circle contained in $H$ and use Fubini: the mean value property follows.
The function $U$ is also bounded, because $u$ is: 
$$
|U(x+iy)|\le 
 \frac{\sup |u|}{\pi} \int_{-\infty}^{\infty}{\frac{y\,  dt}{(t - x)^2 + y^2}} = \sup| u|
$$
Finally, $U(x+iy)\to u(s)$ as $x+iy\to s\in\mathbb R$. Indeed, 
$$
U(x + iy)-u(s) = \frac{1}{\pi} \int_{-\infty}^{\infty}{\frac{y \cdot (u(t)-u(s)) dt}{(t - x)^2 + y^2}}
$$
where the integral is small by a standard argument: when $|t- s|<\delta$, we have $|u(t)-u(s)|<\epsilon$; the integral over $|t-s|>\delta$ tends to $0$ as $y\to 0$. 
So, the difference $u-U$ is harmonic in $H$ and vanishes on the real line. You can argue in various ways that such a function must be identically zero. 
A: Another way to derive the solution is to use the Fourier transform (FT) in x, applying it to u(x,y) to obtain a function w(k,y). Laplace's equation for u becomes a 2nd order linear ODE in y for w  which is easily solved, and the condition that u is bounded in the upper half-plane implies that w(k,y)= G(k) exp(-|k| y), where G(k) is the FT of the function F(x)=u(x,0), the function on the boundary y=0. Then taking the inverse FT, u(x,y) is given by the convolution of F(x) and the inverse FT of the function exp(-|k| y), which is precisely the kernel appearing in the Poisson formula.
