= Radial Fourier representation of the Helmholtz Green function
{title2=$G_+(l)=\frac1{2\pi^2l}\int_0^\infty\frac{k\sin(kl)}{k^2-k_0^2-i0}\,dk$}
For $k_0>0$, $l=|r-r'|>0$ 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 $r-r'$ 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 $G_{\rm PV}(l)=\cos(k_0l)/(4\pi l)$. The outgoing prescription is $k^2-k_0^2-i0$, yielding $G_+(l)=e^{ik_0l}/(4\pi l)$; the opposite prescription gives the incoming exponential. The <Sokhotski–Plemelj formula> $1/(q-i0)=\operatorname{PV}(1/q)+i\pi\delta(q)$ shows that the additional radial contribution is $i\sin(k_0l)/(4\pi l)$.
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 $\Phi=H*V=-G*V$, 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.
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