# Non-vanishing bondary terms of the inhomogenous wave equation

I'm trying to follow a derivation of the solution to the inhomogeneous wave equation $$\bigg[ \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \bigg] \psi(\vec{r},t) = - f(\vec{r},t),$$ which is given by (Andrew Zangwill, Modern Electrodynamics): $$\psi(\vec{r},t) = \int_{t_1}^{t_2} \text{d}t' \int_V \text{d}^3r' G(\vec{r},t|\vec{r}',t')f(\vec{r}',t') \\ + \int_{t_1}^{t_2} \text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot[G(\vec{r},t|\vec{r}',t')\nabla'\psi(\vec{r}',t') - \nabla' G(\vec{r},t|\vec{r}',t')\psi(\vec{r}',t')] \\ + \frac{1}{c^2} \int_V \text{d}^3r' \bigg[\frac{\partial}{\partial t}G(\vec{r},t|\vec{r}',t') \psi(\vec{r}',t') - G(\vec{r},t|\vec{r}',t') \frac{\partial}{\partial t} \psi(\vec{r}',t')\bigg]_{t'=t_1}^{t'=t_2},$$ satisfied by either the advanced or retarded Green function, $$G_\pm (\vec{r},t|\vec{r}',t') = \delta(t - t' \pm |\vec{r} - \vec{r}'|/c) / 4 \pi |\vec{r} - \vec{r}'|.$$ The author identifies the second line of the solution as a spatial boundary term, and the third line as a temporal boundary term. He then asserts that as the spatial boundary $$\partial V$$ is pushed to infinity, the spatial term goes to zero. That said, I can't find any good mathematical reason to disregard this surface integral. Letting $$\vec{R} := \vec{r} - \vec{r}'$$ (for simplicity) and expanding the integrand a bit, $$\int_{t_1}^{t_2} \text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot[G(\vec{r},t|\vec{r}',t')\nabla'\psi(\vec{r}',t') - \nabla' G(\vec{r},t|\vec{r}',t')\psi(\vec{r}',t')] = \\ \frac{1}{4 \pi} \int_{\partial V} \text{d}\vec{S}' \cdot \frac{\nabla'\psi(\vec{r}',t \pm R/c)}{R} - \int_{t_1}^{t_2}\text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot \nabla' G(\vec{r},t|\vec{r}',t')\psi(\vec{r}',t'),$$ where the second integral here can be expanded further using using the chain rule, $$\nabla'G = \frac{\delta(t - t' \pm R/c)}{4 \pi} \nabla' \bigg( \frac{1}{R} \bigg) + \frac{1}{4 \pi R} \nabla'\delta(t - t' \pm R/c).$$ I then shift the gradient on the delta onto $$\psi / R$$ via integration by parts, which allows for some very convenient cancellation of $$\nabla'(1 / R)$$ terms, $$\int_{t_1}^{t_2}\text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot \nabla' G(\vec{r},t|\vec{r}',t')\psi(\vec{r}',t') = \frac{1}{4 \pi} \int_{t_1}^{t_2}\text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot \nabla' \bigg(\frac{1}{R}\bigg)\delta(t - t' \pm R/c) \\ - \frac{1}{4 \pi} \int_{t_1}^{t_2}\text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot \nabla'\bigg(\frac{\psi(\vec{r}',t')}{R} \bigg) \delta(t - t' \pm R/c) \\ = - \frac{1}{4 \pi} \int_{\partial V} \text{d}\vec{S}' \cdot \frac{\nabla'\psi(\vec{r}',t \pm R/c)}{R}$$ Which, to my dismay, would yield a final spatial boundary that is identically zero after adding everything up, were it not for the negative sign on the result of the line just above. As such, I'm left with a final spatial boundary that has a factor of $$1 / 2 \pi$$, $$\int_{t_1}^{t_2} \text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot[G(\vec{r},t|\vec{r}',t')\nabla'\psi(\vec{r}',t') - \nabla' G(\vec{r},t|\vec{r}',t')\psi(\vec{r}',t')] = \\ \frac{1}{2 \pi} \int_{\partial V} \text{d}\vec{S}' \cdot \frac{\nabla'\psi(\vec{r}',t \pm R/c)}{R}$$ And I can't convince myself that this last integral should be zero. If $$\nabla'\psi$$ goes as $$1 / R$$ (as is often the case with radiation fields in electrodynamics), the solid angle form will carry a $$R^2$$ factor, yielding a constant integral. My hope was that the $$G \nabla'\psi$$ and $$\psi \nabla'G$$ integrals would cancel for this form of Green function, but this is ruined by a singular, irritating, lingering negative sign that makes me wonder if I simply made a mistake in my calculation somewhere.

I ask because this spatial boundary term is disregarded by most authors on physical grounds for the retarded solution, as a causal universe forbids the wave from ever reaching the infinite boundary. Yet, for the advanced solution, I have no intuitive sense for why the surface integral should vanish, and must rely on mathematics alone. All sources seem to indicate that this integral disappears for any linear combination of advanced and retarded solutions, as long as the boundary is at infinity. What's the last little step I'm missing?