= Solution
Choose the <Green function> convention $\Delta_{\mathbf r}G(\mathbf r;\mathbf r_0)=\delta^{(3)}(\mathbf r-\mathbf r_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,
$$
\boxed{\Phi(\mathbf r_0)=\int_{\mathbb R^3}G(\mathbf r;\mathbf r_0)\sigma(\mathbf r)\,d^3r,\qquad G=-\frac1{4\pi|\mathbf r-\mathbf r_0|}.}
$$
Away from its source, this <fundamental solution of the Laplace equation> is harmonic. Its radial derivative is $1/(4\pi\rho^2)$, so the flux of its <gradient> through a small source-centred sphere is one. The <divergence theorem> therefore supplies exactly the unit <Dirac delta function> required by the equation. Decay at infinity excludes an added globally regular harmonic term.
For the half-space, reflect the source to $\mathbf r_0^*=(-x_0,y_0,z_0)$. The <method of images> gives the zero-boundary <Dirichlet Green function>
$$
\widetilde G=-\frac1{4\pi}\left(\frac1{|\mathbf r-\mathbf r_0|}-\frac1{|\mathbf r-\mathbf r_0^*|}\right).
$$
The two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.
The <sign conventions for a Dirichlet Green function> require care here. With $\Delta\widetilde G=+\delta$, <Green's second identity> gives $\Phi(\mathbf r_0)=\int_{\partial\Omega}f\,\partial_n\widetilde G\,dS$. At this wall the outward normal is $-\mathbf e_x$, so $\partial_n\widetilde G=-\partial_x\widetilde G$. Direct differentiation gives $\partial_x\widetilde G|_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0)^2+(z-z_0)^2$. Thus the printed intermediate formula with $+\partial_x\widetilde G$ needs a minus sign for its stated Green convention. The final <Poisson kernel for the upper half-space> is positive:
$$
\boxed{\Phi(\mathbf r_0)=\frac{x_0}{2\pi}\int_{\mathbb R^2}\frac{f(y,z)\,dy\,dz}{[x_0^2+(y-y_0)^2+(z-z_0)^2]^{3/2}}.}
$$
Its boundary integral is one for constant data, providing an independent sign check.
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