${\displaystyle \int _{h.s.}[P({\textbf {r}}',{\textbf {r}}_{s},\omega )\nabla 'G_{0}^{+}({\textbf {r}}',{\textbf {r}},\omega )-G_{0}^{+}({\textbf {r}}',{\textbf {r}},\omega )\nabla 'P({\textbf {r}}',{\textbf {r}}_{s},\omega )]\cdot {\hat {n'}}dS'\rightarrow 0}$

as ${\displaystyle r'\rightarrow 0}$. ${\displaystyle h.s.}$ is the upper surface of of closed surface ${\displaystyle S}$, and ${\displaystyle r'=|{\textbf {r}}'-{\textbf {0}}|}$ is its radius.

### Prove

The gradient of 3D causal free space Green function in spherical coordinate has the following form, ${\displaystyle \nabla 'G_{0}^{+}({\textbf {r}}',{\textbf {r}},\omega )={\frac {\partial G_{0}^{+}}{\partial r'}}{\hat {r'}}+{\frac {1}{r'}}{\frac {\partial G_{0}^{+}}{\partial \theta '}}{\hat {\theta '}}+{\frac {1}{r'sin\theta '}}{\frac {\partial G_{0}^{+}}{\partial \phi '}}{\hat {\phi '}}}$

where ${\displaystyle r'}$ is the radial distance from surface element ${\displaystyle dS'}$ to the origin, ${\displaystyle \theta '}$ is the depression angle of ${\displaystyle dS'}$ and ${\displaystyle \phi '}$ is its azimuthal angle. As ${\displaystyle r'\rightarrow \infty }$, both ${\displaystyle {\textbf {r}}_{s}}$ and ${\displaystyle {\textbf {r}}}$ infinitely approaches the origin ${\displaystyle {\textbf {0}}}$ from the view of far field and thus at far field the outgoing reference wave,

$$G_0^+(\textbf{r}',\textbf{r},\omega)=-\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'-\textbf{r}|}}{|\textbf{r}'-\textbf{r}|} \rightarrow -\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'-\textbf{0}|}}{|\textbf{r}'-\textbf{0}|}=-\frac{1}{4\pi}\frac{e^{ikr'}}{r'}=o(\frac{1}{r'})$$

So at infinity, the inner product of ${\displaystyle \nabla 'G_{0}^{+}}$ and normal unit vector of ${\displaystyle dS'}$ becomes

$$\nabla' G_0^+ \cdot \hat{n}'=\nabla ' G_0^+ \cdot \hat{r'}=\frac{\partial G_0^+}{\partial r'}=-\frac{1}{4\pi}(ik\frac{e^{ikr'}}{r'}-\frac{e^{ikr'}}{r'^2})=ikG_0^++o(\frac{1}{r'^2})$$

Meanwhile, according to Lippmann-Schwinger equation, the field anywhere is

$$P(\textbf{r}',\omega)=\int_\infty \rho(\textbf{r}_s')G_0^+(\textbf{r}',\textbf{r}_s',\omega)d\textbf{r}_s'=\int_\infty \rho(\textbf{r}_s')(-\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'-\textbf{r}_s'|}}{|\textbf{r}'-\textbf{r}_s'|})d\textbf{r}_s'$$

where ${\displaystyle \rho ({\textbf {r}}_{s}')}$ is distribution of all sources including airguns and passive sources due to medium perturbation of air and earth. Since at infinitely far field, all sources' location ${\displaystyle {\textbf {r}}_{s}'\rightarrow {\textbf {0}}}$ and all sources are in a finite region, ${\displaystyle \int _{\infty }\rho ({\textbf {r}}_{s}')d{\textbf {r}}_{s}'}$ is finite. So

$$P(\textbf{r}',\omega)\rightarrow \int_\infty \rho(\textbf{r}_s')(-\frac{1}{4\pi}\frac{e^{ik|\textbf{r}'|}}{|\textbf{r}'|})d\textbf{r}_s'=(-\frac{1}{4\pi}\frac{e^{ikr'}}{r'}) \int_\infty \rho(\textbf{r}_s')d\textbf{r}_s'=o(\frac{1}{r'})$$

So

$$\nabla ' P(\textbf{r}',\omega)\cdot \hat{n'}\\ \nonumber =\nabla ' P(\textbf{r}',\omega)\cdot \hat{r'}\\ \nonumber =\int_\infty \rho(\textbf{r}_s')(-\frac{1}{4\pi}\frac{\partial}{\partial r'}\frac{e^{ik|\textbf{r}'-\textbf{r}_s'|}}{|\textbf{r}'-\textbf{r}_s'|})d\textbf{r}_s'\\ \nonumber \rightarrow \int_\infty \rho(\textbf{r}_s')[ikG_0^+(\textbf{r}',\textbf{r}_s',\omega)+o(\frac{1}{r'^2})]d\textbf{r}_s'\\ \nonumber =ikP(\textbf{r}',\omega)+o(\frac{1}{r'^2})\int_\infty \rho(\textbf{r}_s')d\textbf{r}_s'\\ \nonumber =ikP(\textbf{r}',\omega)+o(\frac{1}{r'^2})$$

Therefore, at infinitely far field,

\begin{align} &\int_{h.s.} [P(\textbf{r}',\textbf{r}_s,\omega)\nabla ' G_0^{+}(\textbf{r}',\textbf{r},\omega)-G_0^{+}(\textbf{r}',\textbf{r},\omega)\nabla ' P(\textbf{r}',\textbf{r}_s,\omega)] \cdot \hat{n}' dS'\\ \nonumber &=\int_{h.s.} [P(ikG_0^++o(\frac{1}{r'^2}))-G_0^+(ikP+o(\frac{1}{r'^2}))]dS' \\ \nonumber &=\int_{h.s.} o(\frac{1}{r'^3}) dS' \end{align}

This means the integrand has the order of $r'^{-3}$. Because the semiphere surface has the order of $r'^2$, ultimately the integration has the order of $r'^{-1}$ and vanishes as $r'\rightarrow \infty$.