The formula for half space solutions to the Laplace equation can be given:
\begin{equation}
u(x,y) = \int_{z\in \mathbb{R}} \frac{1}{\pi}\frac{y}{(x-z)^2+y^2}f(z)
\end{equation}
Use this formula to solve the initial value problem:
\begin{equation}
\nabla^2u=0. \quad u(x,0)=f(x)=\frac{1}{1+x^2}
\end{equation}
Now I have gotten the integral into the form (as a hint in the question asks):
\begin{equation}
\int_{z\in \mathbb{R}}\frac{a(z-x)+b}{(z-x)^2+y^2} + \frac{cz+d}{1+z^2}
\end{equation}
Where a,b,c,d do not depend on z.
But have been unable to compute this integral. Wolfram Alpha says the integral diverges. The integral can be Lebesgue integration. Any hints/similar eamples would be great,thanks.
Solve $$u_{xx}+u_{yy}=0, \;-\infty<x<\infty,\; y>0,$$ $$u(x,0)=\frac{1}{1+x^2},\quad u_y(x,0)=0,\;-\infty<x<\infty.$$ I get $$u=\frac{{{x}^{2}}-{{y}^{2}}+1}{\left( {{y}^{2}}-2 y+{{x}^{2}}+1\right) \, \left( {{y}^{2}}+2 y+{{x}^{2}}+1\right) }$$ Solving hint: General solution of Laplace equation is $$u=F(x+iy)+G(x-iy)$$ Question: How depends the solution from condition on $u_y(x,0)$ ?
For example if $u_y(x,0)=1$, solution is $$u=\frac{{{y}^{5}}+2 {{x}^{2}}\, {{y}^{3}}-2 {{y}^{3}}-{{y}^{2}}+{{x}^{4}} y+2 {{x}^{2}} y+y+{{x}^{2}}+1}{\left( {{y}^{2}}-2 y+{{x}^{2}}+1\right) \, \left( {{y}^{2}}+2 y+{{x}^{2}}+1\right) }$$