= Solution
Since $\epsilon$ is per unit mass, each shell contributes $dL=\epsilon\,dm=4\pi r^2\rho\epsilon\,dr$. Both factors of <mass density> must be retained. Substituting the <linear-density stellar model> and its <temperature> profile gives
$$
L(r)=4\pi\epsilon_0\rho_c^2R^3\left(\frac{T_c}{T_0}\right)^\nu\int_0^{r/R}x^2(1-x)^{2+\nu}\left(1+2x-\frac95x^2\right)^\nu\,dx.
$$
Writing $I_\nu(z)$ for this incomplete integral, with $I_\nu=I_\nu(1)$, the <luminosity> is
$$
\boxed{L(r)=\frac{36\epsilon_0M^2}{\pi R^3}\left(\frac{5GM}{12\mathcal R RT_0}\right)^\nu I_\nu(r/R),\qquad L(R)=\frac{36\epsilon_0M^2}{\pi R^3}\left(\frac{5GM}{12\mathcal R RT_0}\right)^\nu I_\nu.}
$$
This 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 $L(R)=6\epsilon_0M^2/(5\pi R^3)$ for temperature-independent heating. This is the generated <luminosity>; the radiative-transport incompatibility noted above is not removed by integrating the heating.
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