= Solution
Integrating the <standard thin-disk dissipation flux> over annular area, with both faces already included in $\mathcal H$, gives
$$
L_D=\frac{3GM_*\dot M}{2}\int_{r_*}^{r_{\rm out}}\left(r^{-2}-r_*^{1/2}r^{-5/2}\right)dr
=\frac{GM_*\dot M}{2r_*}\left[1-3\frac{r_*}{r_{\rm out}}+2\left(\frac{r_*}{r_{\rm out}}\right)^{3/2}\right].
$$
Thus \b[the disk <luminosity>] for $r_{\rm out}\gg r_*$ is
$$
\boxed{L_D\simeq\frac{GM_*\dot M}{2r_*}.}
$$
The <Newtonian gravitational potential> decreases by approximately $GM_*/r_*$ per unit mass from a distant outer edge to the surface, giving a potential-energy release rate $GM_*\dot M/r_*$. The <standard thin-disk luminosity> is half of this.
The missing half remains as <kinetic energy> of nearly circular orbital motion: at the inner edge $v_K^2=GM_*/r_*$, so the specific orbital <kinetic energy> is $GM_*/(2r_*)$. Equivalently, circular motion has total specific <mechanical energy> $-GM_*/(2r_*)$. Matter joining a slowly rotating star must shed this orbital motion in an <accretion-disk boundary layer>, producing approximately another $GM_*\dot M/(2r_*)$ of <luminosity>. A rotating star can retain some energy in spin, so equal disk and <accretion-disk boundary layer> luminosities assume slow stellar rotation. For a central <black hole>, there is no material surface, and energy can instead be carried inward.
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