Consider a domain $D : {(x,y) : x>0 , y>0}$
Let $\mathbf{x}= (x,y)$ and $\mathbf{\xi}= (\xi_x, \xi_y)$. Then the Green's function satisfying $$\nabla^2G = \delta(\mathbf{x} - \mathbf{\xi} )$$
subject to
$$\frac{\partial G}{\partial x} (0, y, \mathbf{\xi})= 0 \, \, \,\,\, \mathrm{for} \, \, \,\,\, y>0$$
$$G(x, 0, \mathbf{\xi}) = 0 \, \, \,\,\, \mathrm{for} \, \, \,\,\, x>0$$
is given by:
$$\frac{1}{4 \pi}\ln \huge| \small\frac{((x - \xi_x)^2 + (y- \xi_y)^2)((x + \xi_x)^2 + (y- \xi_y)^2)}{((x - \xi_x)^2 + (y+ \xi_y)^2)((x + \xi_x)^2 + (y+ \xi_y)^2)} \huge|$$
through the method of images. Now using this I have to solve Laplace's equation:
$$\nabla^2u=0$$ subject to $u \rightarrow 0$ as $|\mathbf{x}| \rightarrow \infty$,
$$\frac{\partial u(0,y)}{\partial x} = h(y) \, \, \, \, \mathrm{for} \, \, \, \, y > 0$$ and $$u(x, 0) = g(x) \, \, \, \, \mathrm{for} \, \, \, \, x > 0$$
So far I've got that
$$G(0, y, \mathbf{\xi}) = \frac{1}{2\pi} \ln \huge| \small \frac{\xi_x^2+(y- \xi_y)^2}{\xi_x^2 +(y+ \xi_y)^2} \huge|$$
which gets me
$$u(\xi_x, \xi_y) = \frac{1}{2\pi} \int^{\infty}_{0}h(y) \ln \huge| \small \frac{\xi_x^2 +(y- \xi_y)^2}{\xi_x^2 +(y+ \xi_y)^2} \huge| \small dy$$
I'm not sure if I'm going about the rest of the question correctly, if I rewrite the above to
$$u(x, y) = \frac{1}{2\pi} \int^{\infty}_{0}h(\lambda) \ln \huge| \small \frac{x^2 +(y- \lambda)^2}{x^2 +(y+ \lambda)^2} \huge| \small d\lambda \, \, \, \,$$
would that be correct?