 Solution

Since $ϵ$ is per unit mass, each shell contributes $dL=ϵdm=4\pi r^2\rho ϵdr$. Both factors of mass density must be retained. Substituting the linear-density stellar model and its temperature profile givesWriting $I_{\nu }(z)$ for this incomplete integral, with $I_{\nu }=I_{\nu }(1)$, the luminosity isThis supplies both the local $L_r$ and the total luminosity associated with the printed integral extending to one. For fixed composition and fixed $\nu$, the total scales as $M^{2+\nu }{R}^{-3-\nu }$. The physical nuclear-heating exponents are positive; mathematically the surface integral converges for $\nu >-3$. As a simple normalization check, $I_0=1/30$, hence $L(R)=6{ϵ}_0{M}^2/(5\pi R^3)$ for temperature-independent heating. This is the generated luminosity; the radiative-transport incompatibility noted above is not removed by integrating the heating.