= Solution
For fixed source $(\xi,\eta)\in D$, define the <Dirichlet Green function> by $-\Delta_{x,y}G=\delta_{(\xi,\eta)}$ in $D$, $G=0$ on $\partial D$. It is <harmonic> away from the source and has <singularity> $-(2\pi)^{-1}\log\sqrt{(x-\xi)^2+(y-\eta)^2}$. The outward <normal vector> <derivative> is $\partial_nG=\mathbf n\cdot\nabla_{x,y}G$ at a boundary point of arc-length parameter $s$.
<Green's second identity> gives $\int_D(u\Delta G-G\Delta u)=\int_{\partial D}(u\partial_nG-G\partial_nu)\,ds$. Substituting the two PDEs and the boundary data yields
$$
-u(\xi,\eta)+\int_DfG=\int_{\partial D}g\partial_nG,\qquad
\boxed{u(\xi,\eta)=-\int_{\partial D}g\partial_nG\,ds+\int_DfG\,dx\,dy.}
$$
The singular version follows by excising a small source disk and taking its radius to zero, or directly by the distributional identity.
For the upper half-plane use the opposite-sign image source at $(\xi,-\eta)$:
$$
\boxed{G=\frac1{4\pi}\log\frac{(x-\xi)^2+(y+\eta)^2}{(x-\xi)^2+(y-\eta)^2}.}
$$
The image lies outside the domain, so it is <harmonic> inside; the true source has the required unit <singularity>, and the two distances agree at $y=0$, giving zero boundary value. On that boundary $\mathbf n=-\mathbf e_y$, so
$$
\partial_nG=-\frac\eta{\pi[(s-\xi)^2+\eta^2]}.
$$
The assumed far-field decay makes the large semicircle contribution vanish. With $f=0$, the representation becomes
$$
\boxed{u(\xi,\eta)=\frac\eta\pi\int_{-\infty}^\infty\frac{g(s)}{(\xi-s)^2+\eta^2}\,ds.}
$$
Finally set $s=\xi+\eta\tan\theta$, $ds=\eta\sec^2\theta\,d\theta$. The <Jacobian matrix> cancels the denominator, giving
$$
\boxed{u(\xi,\eta)=\frac1\pi\int_{-\pi/2}^{\pi/2}g(\xi+\eta\tan\theta)\,d\theta.}
$$
If $g$ is even, the substitution $s\mapsto-s$ in the <Poisson integral> gives $u(-\xi,\eta)=u(\xi,\eta)$. At $\eta=0$ this agrees with the continuous even boundary trace.
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