Radial Fourier representation of the Helmholtz Green function ($G_{+}(l)=\frac{1}{2{\pi }^{2}l}{\int }_0^{\infty }\frac{k\sin (kl)}{k^2-k_0^2-i0}dk$)

For $k_0>0$, $l=|r-r^{\prime }|>0$ and the convention $(\Delta +k_0^2)G=-\delta$, the spatial Fourier transform gives the multiplier $(k^2-k_0^2{)}^{-1}$. Aligning the Fourier polar axis with $r-r^{\prime }$ gives ${\int }_{S^2}{e}^{ikl\cos \theta }d\Omega =4\pi \sin (kl)/(kl)$ and hence the displayed radial integral.

The pole on the integration path requires a prescription. The Cauchy principal value gives $G_{\mathrm{PV}}(l)=\cos (k_{0}l)/(4\pi l)$. The outgoing prescription is $k^2-k_0^2-i0$, yielding $G_{+}(l)=e^{i{k}_{0}l}/(4\pi l)$; the opposite prescription gives the incoming exponential. The Sokhotski–Plemelj formula $1/(q-i0)=\mathrm{PV}(1/q)+i\pi \delta (q)$ shows that the additional radial contribution is $i\sin (k_{0}l)/(4\pi l)$.

For the inverse-kernel convention $(\Delta +k_0^2)H=+\delta$, use $H=-G$. Thus a source equation $(\Delta +k_0^2)\Phi =V$ has $\Phi =H\ast V=-G\ast V$, with the same boundary or radiation convention throughout. Without a pole prescription or an equivalent condition at infinity, the real-frequency source problem does not specify a unique Green function.