Solution to Dirichlet boundary value problem on upper halve plane using Green's function Currently I am studying for an exam about partial differential equations. While looking through some of the exercises concerning Green's function on the plane, I came across a rather impenetrable-seeming integral connected to a Dirichlet BVP. As the textbook we are using (Partial Differential Equations, Peter Olver) does not provide clear examples on this topic, I figured that the internet would be a fine next step towards a solution. I also think that some experienced mathematicians out there might enjoy an exercise, and explaining it. Let me be clear: this exercise will not be examined. The problem is as follows:
Solve for $u$ where
\begin{align}
-\Delta u (x,y)= \frac{1}{1+y}\text{ on }\{(x,y):y>0\}\text{ and }u(x,0) = 0\text{ for all }x\in \mathbb{R}.
\end{align}
Now, we are asked to use the Green's function for the upper half plane to find a solution. Using the method of images, one finds such a function quite easily:
\begin{align}
G(x,y;\theta,\eta) = \frac{1}{4\pi}\log\left(\frac{(x-\theta)^2 + (y-\eta)^2}{(x-\theta)^2+(y+\eta)^2}\right).
\end{align}
By Green's representation formula, the solution is then given by
\begin{align}u(x,y) = \frac{1}{4\pi}\int_{-\infty}^{\infty}\int_0^{\infty}\frac{1}{1+\eta}\log\left(\frac{(x-\theta)^2 + (y-\eta)^2}{(x-\theta)^2+(y+\eta)^2}\right)\,\mathrm{d}\eta\,\mathrm{d\theta}.
\end{align}
I am wondering, did the author perhaps choose this example problem because the resulting integral would be solvable? If so, could someone shed light on the problem? And if not, is there a better way to solve it?
My thanks.
 A: I'll assume you made a typo in the equation, since the Green's function you have is for $\nabla$, not $-\nabla$
The inhomogeneous part only depends on $y$, so you can guess a solution of the form $u(x,y) = g(y)$, where
$$ -g''(y) = \frac{1}{1+y}, \ g(0) = 0 $$
Then integrating twice gives
$$ g(y) = (1+y)\ln(1+y) - y + cy $$
where $c$ is some arbitrary constant.

Change the order of integration to.
$$ \frac{1}{4\pi}\int_0^\infty \frac{1}{1+\eta}\int_{-\infty}^\infty \ln \left(\frac{(\theta-x)^2 + (\eta-y)^2}{(\theta-x)^2 + (\eta+y)^2}\right)\ d\theta\ d\eta $$
Substitute $u = \theta-x$ and rewrite $a = |\eta-y|$, $b=|\eta+y|$. The inner integral can be evaluated as
\begin{align} 
\int_{-\infty}^\infty \ln\left(\frac{u^2+a^2}{u^2+b^2}\right) du &= u\ln\left(\frac{u^2+a^2}{u^2+b^2}\right)\Bigg|_{-\infty}^\infty - \int_{-\infty}^\infty \left(\frac{2u^2}{u^2+a^2} - \frac{2u^2}{u^2+b^2}\right)du \\
&= 2\int_{-\infty}^\infty \left(\frac{a^2}{u^2+a^2} - \frac{b^2}{u^2+b^2}\right) du \\
&= 2\left(a\arctan\frac{u}{a} - b\arctan\frac{u}{b}\right)\Bigg|_{\infty}^\infty \\
&= 2\pi (a-b)
\end{align}
So we're left with
\begin{align} 
\frac12 \int_0^\infty \frac{1}{1+\eta}\big(|\eta-y|-|\eta+y|\big) 
d\eta
&= \int_0^y \frac{1}{1+\eta}(-\eta)d\eta + \int_y^\infty \frac{1}{1+\eta}(-y)d\eta \\
&= -\int_0^y \left(1 - \frac{1}{1+\eta}\right)d\eta - y\int_y^\infty \frac{1}{1+\eta}d\eta \\
&= -y + \ln(1+y)+y\ln(1+y) + y\lim_{b\to\infty}\ln(1+b)
\end{align}
You may notice that the last term of the integral does not converge. This is because the solution is not unique. However, if you add a second boundary condition such that $u_y(x,b) = g'(b) = 0$, then the solution is indeed
$$ u(x,y) = (1+y)\ln(1+y) - y + y\ln(1+b), \ y \in (0,b) $$
for any $b$ arbitrarily large. This matches the solution obtained through integration as above.
