Greens function for 2d laplace equation with neumann boundary conditions I have a domain,
 $ D : {(x,y) : x>0 , y>0}$ 
Let $ \mathbf{x}= (x,y) $ and $\mathbf{\xi}= (\xi_x, \xi_y)$,
The Greens function satisfying: 
$$\nabla^2G = \delta(\mathbf{x} - \mathbf{\xi} )$$
with conditons:
$$ G(x, 0, \mathbf{\xi}) = 0 \, \, \,\,\, \mathrm{for} \, \, \,\,\,
x>0$$
$$ \frac{\partial G(0, y, \mathbf{\xi})}{\partial x} = 0 \, \, \,\,\, \mathrm{for} \, \, \,\,\,
y>0$$
(using the method of images) is found to be: 
$$ \frac{1}{4 \pi}\ln \huge| \small\frac{((x - \xi_x)^2 + (y- \xi_y)^2)((x + \xi_x)^2 + (y- \xi_y)^2)}{((x - \xi_x)^2 + (y+ \xi_y)^2)((x + \xi_x)^2 + (y+ \xi_y)^2)} \huge|$$
(please correct me if i have miscalculated). I now want to solve Laplace's equation:
$$\nabla^2u=0$$
such that $u \rightarrow 0 $ as $|\mathbf{x}| \rightarrow \infty$ and:
$$ u(x, 0) = g(x) \, \, \, \, \mathrm{for} \, \, \, \, x > 0 $$
$$ \frac{\partial u(0,y)}{\partial x} = h(y)  \, \, \, \, \mathrm{for} \, \, \, \, y > 0 $$
I have obtained that 
$$G(0, y, \mathbf{\xi}) = \frac{1}{2\pi} \ln \huge| \small \frac{\xi_x^2+(y- \xi_y)^2}{\xi_x^2 +(y+ \xi_y)^2} \huge|$$
So this is where i am not so sure i am correct, we have:
$$u(\xi_x, \xi_y) = \frac{1}{2\pi} \int^{\infty}_{0}h(y) \ln \huge| \small \frac{\xi_x^2 +(y- \xi_y)^2}{\xi_x^2 +(y+ \xi_y)^2} \huge| \small dy$$
or equivalently:
$$u(x, y) = \frac{1}{2\pi} \int^{\infty}_{0}h(\lambda) \ln \huge| \small \frac{x^2 +(y- \lambda)^2}{x^2 +(y+ \lambda)^2} \huge|  \small d\lambda \, \, \, \, (*)$$
Which yields:
$$u(x,0) = \frac{1}{2\pi} \int^{\infty}_{0}h(\lambda) \ln \huge| \small \frac{x^2 +\lambda^2}{x^2 + \lambda^2} \huge| \small d\lambda \normalsize = g(x) = 0$$
Is my answer for $g(x)$ correct? If so, do i have $u(x,y)$ in the simplest form possible?
 A: We present an outline only for obtaining the solution to the posed problem for $u$ and provide an outline of a way for verifying the solution.  
We proceed by using Green's Identity to write
$$\int_S (u(\vec r')\nabla'^2G(\vec r,\vec r')-G(\vec r,\vec r')\nabla'^2u(\vec r'))\,dS'=\oint_C (u(\vec r')\nabla' G(\vec x,\vec r')-G(\vec r,\vec r')\nabla' u(\vec r'))\cdot \hat n'\,d\ell' \tag 1$$
Using $\nabla' ^2G(\vec r,\vec r')=\delta(\vec r-\vec r')$ and $\nabla' ^2u(\vec r')=0$, the left-hand side of $(1)$ becomes
$$u(\vec r)=\int_S (u(\vec r')\nabla'^2G(\vec r,\vec r')-G(\vec r,\vec r')\nabla'^2u(\vec r'))\,dS' \tag 2$$
Using the boundary conditions, 
$$\begin{align}
u(x,0)&=g(x)\\\\
G(x,0)&=0\\\\
\left.\frac{\partial u(x,y)}{\partial x}\right|_{x=0}&=h(y)\\\\
\left.\frac{\partial G(x,y;x',y')}{\partial x'}\right|_{x'=0}&=0\\\\
\end{align}$$
the integral on the right-hand side of $(1)$ can be written
$$\begin{align}
\oint_C (u(\vec r')\nabla' G(\vec x,\vec r')-G(\vec r,\vec r')\nabla' u(\vec r'))\cdot \hat n'\,d\ell'&=\int_0^\infty g(x')\left.\frac{\partial G(x,y;x',y')}{\partial y'}\right|_{y'=0}\,dx'\\\\
&-\int_0^\infty h(y')G(x,y;0,y')\,dy' \tag 3
\end{align}$$
Equating $(2)$ and $(3)$ yields the solution for $u(x,y)$ as
$$\bbox[5px,border:2px solid #C0A000]{u(x,y)=\int_0^\infty g(x')\left.\frac{\partial G(x,y;x',y')}{\partial y'}\right|_{y'=0}\,dx'-\int_0^\infty h(y')G(x,y;0,y')\,dy'} \tag 4$$

To check if $(4)$ is indeed a solution, we need to ensure that $u$ satisfies Laplace's equation for $x>0$, $y>0$ and the prescribed boundary conditions.  
Since $G$ satisfies Laplace's equation for $(x,y)\ne (x',y')$, then $u$ as given in $(4)$ satisfies Laplace's equation.  
Next, we examine $\lim_{y\to 0^+}u(x,y)$.  Using heuristic formal analysis, we have for "small" $\delta >0$ 
$$\begin{align}
\lim_{y\to 0^+} u(x,y)&=\lim_{y\to 0^+}\int_0^\infty g(x')\left.\frac{\partial G(x,y;x',y')}{\partial y'}\right|_{y'=0}\,dx'\\\\
&=\frac{1}{\pi}\lim_{y\to 0^+}\int_0^\infty g(x')\left(\frac{y}{(x-x')^2+y^2}\right)\,dx'\\\\
&\sim \frac{1}{\pi} g(x)\lim_{y\to 0^+}\int_{-\delta}^\delta \frac{y}{x^2+y^2}\,dx\\\\
&=\frac1\pi g(x) \lim_{y\to 0^+}\left( \left.\arctan(x/y)\right|_{-\delta}^{\delta}\right)\\\\
&=g(x)
\end{align}$$
as expected.
Lastly, we examine $\lim_{x\to 0^+}u(x,y)$.  Again, using heuristic formal analysis, we have for "small" $\delta >0$
$$\begin{align}
\lim_{x\to 0^+} \frac{\partial u(x,y)}{\partial x}&=-\lim_{x\to 0^+}\int_0^\infty h(y')\frac{\partial G(x,y;0,y')}{\partial x}\,dy'\\\\
&=\frac{1}{\pi}\lim_{x\to 0^+}\int_0^\infty h(y')\left(\frac{x}{x^2+(y-y')^2}\right)\,dy'\\\\
&\sim \frac{1}{\pi} h(y)\lim_{x\to 0^+} \int_{-\delta}^\delta \frac{y}{x^2+y^2}\,dx\\\\
&=h(y)
\end{align}$$
as expected!
