= Solution
The <outgoing Green function for the three-dimensional Helmholtz equation> satisfies
$$
(\Delta_y+\omega^2)G_F(x,y)=-\delta^{(3)}(y-x).
$$
Its normalization follows by integrating over a small ball about $x$: $G_F\sim1/(4\pi|y-x|)$, so its outward derivative has flux $-1$. Apply <Green's second identity> in the $y$ variables:
$$
\begin{aligned}
\int_{\partial V}(\Psi\nabla_yG_F-G_F\nabla_y\Psi)\cdot n\,dS
&=\int_V(\Psi\Delta_yG_F-G_F\Delta_y\Psi)dV_y\\
&=-\int_V\Psi(y)\delta^{(3)}(y-x)dV_y.
\end{aligned}
$$
The $\omega^2$ terms cancel. Thus the surface integral equals $-\Psi(x)$ for $x\in V$, and zero for $x$ outside the closed volume. Equivalently, remove a small source ball and evaluate its flux before taking its radius to zero. A source on the boundary would require a different limiting formula and is not part of this calculation.
For the upper half-space, take a half-ball of radius $\rho$. Interpret the first far-field condition as $\Psi=O(\rho^{-1})$, and the radiation limit in its usual direction-uniform sense, namely the <Sommerfeld radiation condition>
$$
(\partial_\rho-i\omega)\Psi=o(\rho^{-1}).
$$
This is the finite-bound interpretation of the PDF's first condition. For fixed $x$ and real $\omega$,
$$
G_F=O(\rho^{-1}),\qquad(\partial_\rho-i\omega)G_F=O(\rho^{-2}).
$$
Indeed $|y-x|=\rho-\widehat y\cdot x+O(\rho^{-1})$, and its radial derivative is $1+O(\rho^{-2})$. On the large hemisphere the integrand can therefore be rewritten as
$$
\Psi(\partial_\rho-i\omega)G_F-G_F(\partial_\rho-i\omega)\Psi
=O(\rho^{-3})+o(\rho^{-2}).
$$
Its area is $2\pi\rho^2$, so its integral tends to zero. The planar integral, taken over expanding disks if needed, consequently equals $-\Psi(x)$ for $x_3>0$ and zero for $x_3<0$.
The outward <unit normal vector> on the bottom plane is $-e_3$. Denote this planar integral by
$$
I(x)=\int_{\mathbb R^2}\left[-\Psi(y',0)\partial_{y_3}G_F(x,(y',0))+G_F(x,(y',0))\partial_3\Psi(y',0)\right]d^2y'.
$$
For $x_3>0$ reflect the source point to $x^*=(x_1,x_2,-x_3)$. On the plane,
$$
G_F(x^*,y)=G_F(x,y),\qquad \partial_{y_3}G_F(x^*,y)=-\partial_{y_3}G_F(x,y).
$$
Adding $I(x)=-\Psi(x)$ and $I(x^*)=0$ cancels the boundary values of $\Psi$ and gives $2\int G_F\partial_3\Psi=-\Psi(x)$. Thus the <Neumann representation for a radiating Helmholtz half-space> is
$$
\boxed{\Psi(x)=-\frac1{2\pi}\int_{\mathbb R^2}\partial_3\Psi(y',0)\frac{e^{i\omega\sqrt{|x'-y'|^2+x_3^2}}}{\sqrt{|x'-y'|^2+x_3^2}}\,d^2y',\qquad x_3>0.}
$$
The negative sign is consistent with using the upward derivative $\partial_3\Psi$ rather than the outward <normal derivative>. If that upward derivative is zero everywhere on the plane, the integral is zero. Hence \b[$\Psi=0$ throughout the upper half-space] in the specified radiating class; taking a difference also gives uniqueness for prescribed <Neumann boundary data>.
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