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.