I'm looking for a proper interpretation of the following, which is a variant of the Olbers' paradox.
Consider a spatially closed empty universe ($k = 1$) with a positive cosmological constant $\Lambda$ (closed deSitter universe). It is easy to solve the Friedmann-LemaƮtre-Robertson-Walker equations to find the scale factor : $$\tag{1} a(t) = \ell_{\Lambda} \cosh{(t/\ell_{\Lambda})}, $$ where $\ell_{\Lambda} = \sqrt{\frac{3}{\Lambda}}$. The cosmological time doesn't have any boundary : $-\, \infty < t < \infty$ (no Big Bang, no Big Crunch). Yet, because of the expansion of space, there are two horizons, of the following proper distance ($t_0$ is the present time of the stationnary observer) : \begin{align} \mathcal{D}_{P}(t_0) &= a( t_0) \int_{-\, \infty}^{t_0} \frac{1}{a(t)} \; dt \\[12pt] &= \ell_{\Lambda} \arccos{\big(- \tanh{(t_0 / \ell_{\Lambda})} \big)} \cosh{(t_0 / \ell_{\Lambda})}, \tag{2} \\[12pt] \mathcal{D}_{E}(t_0) &= a( t_0) \int_{t_0}^{\infty} \frac{1}{a(t)} \; dt \\[12pt] &= \ell_{\Lambda} \Big( \pi - \arccos{\big(- \tanh{(t_0 / \ell_{\Lambda})} \big)} \Big) \cosh{(t_0 / \ell_{\Lambda})}. \tag{3} \end{align} The first distance represents the particle horizon, while the second distance represents the event horizon of that empty deSitter spacetime.
Now, it can be proved that the total apparent luminosity at time $t_0$ of all the "fake stars" uniformly distributed in any isotropic/homogeneous space is given by the following integral : $$\tag{4} \mathcal{I}(t_0) = \int I \, n \; d^3 x = L_{\text{e}} \, n_0 \int_{t_{\text{min}}}^{t_0} \frac{a(t)}{a(t_0)} \; dt, $$ where $L_{\text{e}}$ is the absolute luminosity (or power) of a single star, and $n_0$ is the current number density of stars at time $t_0$. In the case of the dust-like matter in an euclidian flat space ($k = 0$, $a(t) \propto t^{2/3}$ and $t_{\text{min}} = 0$), this integral converges and solves the Olbers paradox. But in the case of the empty deSitter space defined above, the scale factor (1) gives (because of $t_{\text{min}} = -\, \infty$) $$\tag{5} \mathcal{I} = \infty. $$ I'm puzzled by this result, since there is still a particle (or causality) horizon, which gives $\mathcal{D}_{P}(t_0) \rightarrow \ell_{\Lambda}$ when $t_0 \rightarrow -\, \infty$ (lower limit of expression (2)), and since the space section is closed (finite number of "fake stars") : $$\tag{6} N_{\text{tot}} = n(t) \, 2 \pi^2 a^3(t) = n_0 \, 2 \pi^2 a^3 (t_0) = cste.$$
So how should we interpret the result of an infinite total apparent luminosity of a finite number of stars in a closed universe, with a particle horizon ?
Note that the space is exponentially contracting during the interval $-\, \infty < t < 0$ (see the scale factor (1)), so I suspect this is a clue to the proper interpretation (the lower limit $t_{\text{min}} = -\, \infty$ is blowing up the integral (4)).
Also, note that the maximal instantaneous proper distance in the closed space is $\mathcal{D}_{\text{max}}(t) = \pi \, a(t)$.
I'm not sure anymore that I should use $t_{\text{min}} = -\, \infty$ in the integral (4) above. This may be a clue to the origin of my problem. Light may turn around several times in the closed universe before beeing detected by the observer, so this may produce the infinite luminosity if the stars are emitting light since infinity in the past. But should we take this into account ? Luminosity should be defined to collect all the light that is reaching a given point (the observer) at a given time $t_0$, so several turns around shouldn't be included, I believe. Each time the light is reaching the observer's location at some time, it counts for the luminosity at that time. See my comments below. What do you think ?