Infinite integral that seems to diverge but question requires it does not. 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.
Let
$$\begin{equation}
u(x,y) = \int_{z\in \mathbb{R}} \frac{1}{\pi}\frac{y}{(x-z)^2+y^2}f(z)\\=
\int_{z\in \mathbb{R}}\frac{a(z-x)+b}{(z-x)^2+y^2}dz+
\int_{z\in \mathbb{R}}\frac{cz+d}{1+z^2}dz\\
=I_1+I_2
\end{equation}$$
Free CAS MAXIMA says that principal values of integrals is
$$I_1=\frac{y\, \left( {{y}^{2}}+{{x}^{2}}-1\right) }{\left( {{y}^{2}}-2 y+{{x}^{2}}+1\right) \, \left( {{y}^{2}}+2 y+{{x}^{2}}+1\right) }$$
$$I_2=-\frac{{{y}^{2}}-{{x}^{2}}-1}{\left( {{y}^{2}}-2 y+{{x}^{2}}+1\right) \, \left( {{y}^{2}}+2 y+{{x}^{2}}+1\right) }$$
Then final integral is
$$u=I_1+I_2=\frac{y+1}{{{y}^{2}}+2 y+{{x}^{2}}+1}$$
Solve
$$u_{xx}+u_{yy}=0, \;-\infty&lt;x&lt;\infty,\; y&gt;0,$$
$$u(x,0)=\frac{1}{1+x^2},\quad u_y(x,0)=0,\;-\infty&lt;x&lt;\infty.$$
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
$$u=\frac{{{y}^{5}}+2 {{x}^{2}}\, {{y}^{3}}-2 {{y}^{3}}-{{y}^{2}}+{{x}^{4}} y+2 {{x}^{2}} y+y+{{x}^{2}}+1}{\left( {{y}^{2}}-2 y+{{x}^{2}}+1\right) \, \left( {{y}^{2}}+2 y+{{x}^{2}}+1\right) }$$