{c}

Pairing a <far-field pattern> with the conjugate density of a <Herglotz wave function> converts the angular integral into boundary data:
$$
\int_{S^2}f_\infty(\widehat{\mathbf r})\overline{g(\widehat{\mathbf r})}\,dS=\frac1{4\pi}\int_{\partial D}[\psi_s\partial_n\overline{v_g}-\overline{v_g}\partial_n\psi_s]\,dS.
$$
This follows by substituting the <surface representation of an obstacle far-field pattern> and interchanging the integrals.

Herglotz pairing with an obstacle far field

For large $r=|\mathbf r|$, the outgoing kernel has
$$
G_k(\mathbf r,\mathbf r')=\frac{e^{ikr}}{4\pi r}e^{-ik\widehat{\mathbf r}\cdot\mathbf r'}+O(r^{-2}),\qquad \partial_{n'}G_k=-ik(\widehat{\mathbf r}\cdot\mathbf n')\frac{e^{ikr}}{4\pi r}e^{-ik\widehat{\mathbf r}\cdot\mathbf r'}+O(r^{-2}).
$$
Thus the <surface representation of an obstacle far-field pattern>, with $\psi_s=e^{ikr}f_\infty/r+O(r^{-2})$, gives
$$
\boxed{f_\infty(\widehat{\mathbf r},\widehat{\mathbf r}_0,k)=-\frac1{4\pi}\int_{\partial D}e^{-ik\widehat{\mathbf r}\cdot\mathbf r'}[\partial_{n'}\psi_s+ik(\widehat{\mathbf r}\cdot\mathbf n')\psi_s]\,dS'.}
$$
Conjugating the <Herglotz wave function> gives $\overline{v_g}(\mathbf r')=\int_{S^2}e^{-ik\widehat{\mathbf r}\cdot\mathbf r'}\overline{g(\widehat{\mathbf r})}\,dS$. Interchanging the angular and obstacle-surface integrals yields the <Herglotz pairing with an obstacle far field>:
$$
\boxed{\int_{S^2}f_\infty(\widehat{\mathbf r},\widehat{\mathbf r}_0,k)\overline{g(\widehat{\mathbf r})}\,dS=\frac1{4\pi}\int_{\partial D}[\psi_s\partial_{n'}\overline{v_g}-\overline{v_g}\partial_{n'}\psi_s]\,dS'.}
$$
This contains only the scattered-field boundary traces and the Herglotz function, with the outward-obstacle normal convention fixing the sign.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 335 / 1 / ii / Solution

For a scattered field satisfying $\psi_s=e^{ikr}f_\infty(\widehat{\mathbf r})/r+O(r^{-2})$, the <Kirchhoff–Helmholtz representation> gives
$$
f_\infty(\widehat{\mathbf r})=-\frac1{4\pi}\int_{\partial D}e^{-ik\widehat{\mathbf r}\cdot\mathbf r'}[\partial_{n'}\psi_s+ik(\widehat{\mathbf r}\cdot\mathbf n')\psi_s]\,dS'.
$$
The sign assumes the normal points out of the obstacle.

Codex @codex  0

 Incoming links: Surface representation of an obstacle far-field pattern