separation of variables. To this end, we note that, by , $dt/d\eta = a$, and therefore $$\int_{- \infty}^{t_{collapse}}  a(t)^{-4} dt = \int_0^{\eta_0} \frac{d\eta}{a(t(\eta))^3}= 8\left(\frac{2GM_0}{R^3_0}\right)^{3/2}\int_0^{\eta_0}\frac{d\eta}{(1+\cos \eta)^3}.$$

Defining then, for $\eta \in [0,\pi)$, $$F(\eta) := \int_0^{\eta}\frac{dx}{(1+\cos x)^3} = \frac{\sin \eta(6\cos \eta+\cos(2\eta)+8)}{15(1+\cos \eta)^3},$$ we obtain, neglecting numerical constants of order 1, $$\label{eq:omegacoll}
\omega(t_{collapse})^2 =  \left[\frac{1}{\omega_0^4}+e^2 \lambda_P^4M \left(\frac{R_0^3}{2GM_0}\right)^{1/2}F(\eta_0)\right]^{-1/2},$$ which is smaller than $\omega_0^2$, as it should.

One can then observe that for $M_0 \to 0$ one has $\eta_0 = \cos^{-1}(4GM_0/R_0 -1) \sim \pi -\sqrt{8GM_0/R_0}$ and therefore $F(\eta_0) \sim \frac85(R_0/8GM_0)^{5/2}$, so that $$\omega(t_{collapse})^2 \sim \frac{G^{3/2}M_0^{3/2}}{e\lambda_P^2 MR_0^2  },$$ would actually be very small with respect to $\omega_0^2$, making the total radiated energy  non negligible.

(Note that for ordinary matter the collapse would stop much before that the matter itself is hidden inside the horizon, due to the non vanishing pressure).

Moreover, this effect might disappear if one takes properly into account the red-shift of the radiation emitted near to the horizon. This could probably be done by using the general relativistic version of the radiated power by a magnetic dipole instead of .

Conversely, for finite values of $M_0$, one can expand  due to the smallness of $\lambda_P^4$, and obtain for the total radiated energy $$\cE 