I set up a basic Laplace's Equation to try solving using Green's Function, and I noticed that the infinite integral diverges. The problem is as follows: $$u_{xx} + u_{yy} = 0; \ \ \ \ \ x>0, y>0 \\ \cases{u(x,0) = -x^2;\ \ \lim_{y\rightarrow\infty} u(x,y) = \infty \\ u(0,y) = y^2;\ \ \lim_{x\rightarrow\infty} u(x,y) = -\infty}$$ The solution is $u(x,y) = y^2-x^2$.
Following This document up to example iii on page 6 we get Green's Function for a upper right quadrant domain PDE for a Laplacian operator. This is obtained via utilizing reflected images of some point source and this gives: $$G(x,y,\zeta,\eta) = \frac{1}{\sqrt{2\pi}}\bigg(\ln\Big(\sqrt{(\zeta-x)^2+(\eta-y)^2}\Big)-\ln\Big(\sqrt{(\zeta-x)^2+(\eta+y)^2}\Big)-\ln\Big(\sqrt{(\zeta+x)^2+(\eta-y)^2}\Big)+\\ \ln\Big(\sqrt{(\zeta+x)^2+(\eta+y)^2}\Big)\bigg)$$ and then plugging in Green's Function into Green's Second Identity gives the solution: $$u(x,y) = \int_Cf\nabla G\cdot \hat{n}dS = \frac{4yx}{\pi}\bigg(-\int_0^\infty\frac{\eta f(0,\eta)}{(x^2+(y+\eta)^2)(x^2+(y-\eta)^2)}d\eta +\int_0^\infty\frac{\zeta f(\zeta,0)}{((x-\zeta)^2+y^2)((x+\zeta)^2+y^2)}d\zeta\bigg)$$ Where we also note from the document on page 2, equation 4, the following definitions: $$\nabla^2u = F \ \ in\ \ D \\ u = f \ \ on \ \ C$$ So plugging in for the solution $u(x,y)$ gives: $$u(x,y) = \frac{4yx}{\pi}\bigg(-\int_0^\infty\frac{\eta \ (u(0,\eta)=\eta^2) = \eta^3}{(x^2+(y+\eta)^2)(x^2+(y-\eta)^2)}d\eta +\int_0^\infty\frac{\zeta \ (f(\zeta,0)=-\zeta^2)=-\zeta^3}{((x-\zeta)^2+y^2)((x+\zeta)^2+y^2)}d\zeta\bigg)$$
The problem is that these integrals for the solution diverge. So I wanna know if this is not the correct Green's Function for the Boundary Value Problem I created even though it uses the Laplacian Operator, or if I made a mistake plugging in for $f$. Or was it something else?