= Solution
Let $m_\chi$ be the <dark matter> particle <mass>, so its number density is $n_\chi=\rho/m_\chi$. Interpret $\epsilon$ as total emitted power per particle, and assume isotropic optically thin <dark-matter decay radiation>; volume emissivity is $j_d=\epsilon\rho/m_\chi$. At projected radius $R$, a line-of-sight coordinate $z$ has $r^2=R^2+z^2$. Writing $A=V^2/(2\pi G)$, the <surface brightness> is
$$
I_d(R)=\frac1{4\pi}\int_{-\infty}^\infty j_d\,dz
=\frac{\epsilon A}{4\pi m_\chi}
\int_{-\infty}^\infty\frac{dz}{R^2+z^2}
=\frac{\epsilon A}{4m_\chi R}.
$$
Here $z=R\tan\theta$ turns the <integral> into $\pi/R$. \b[For the isothermal outer halo,]
$$
\boxed{I_d(R)=\frac{\epsilon V^2}{8\pi Gm_\chi R}\propto R^{-1}.}
$$
If emissivity was defined per unit solid angle, the factor $1/(4\pi)$ is already included in $\epsilon$; this only changes the normalization. The radial law holds between any core and outer truncation. For example, a phenomenological cored profile $\rho=A/(r^2+r_c^2)$ gives $I_d=\epsilon A/[4m_\chi\sqrt{R^2+r_c^2}]$: finite central <surface brightness>, but the same $R^{-1}$ outer law. This cored interpolation is not being asserted to be the exact isotropic constant-dispersion equilibrium.
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