= Solution
Restore the <Fourier transform> integral and the last-scattering distance:
$$
\Theta(\widehat{\mathbf n})=\frac13\int\frac{d^3k}{(2\pi)^3}\Phi_{\mathbf k}
e^{i\mathbf k\cdot\widehat{\mathbf n}\chi_*}.
$$
Insert the <Rayleigh plane-wave expansion> and use <orthogonality> of <spherical harmonics> with the complex conjugate on the projection harmonic. The angular integral selects the matching indices, giving
$$
\boxed{a_{\ell m}=\frac{4\pi i^\ell}{3}\int\frac{d^3k}{(2\pi)^3}
\Phi_{\mathbf k}j_\ell(k\chi_*)Y_{\ell m}^*(\widehat{\mathbf k}).}
$$
The complex conjugation is present in the original PDF. If $A_{\mathbf k}=A\,k^{n/2}$ denotes a mode coefficient with its random amplitude and phase included in $A$, substitution of $\Phi_{\mathbf k}=-6A_{\mathbf k}/k^2$ gives the equivalent factor $-8\pi i^\ell A_{\mathbf k}/k^2$ inside the Fourier integral. The monopole and local observer dipole removed in part (b) do not change this formula for $\ell\geq2$.
Back to article page
