What happens to the Planck distribution if the temperature is set to zero? BE Problem
I am currently working on modelling the density of states and optical conductivity of graphene utilizing the GW algorithm. In calculating the exchange self energy of the system, the formula I am currently using is
$$f(w,T) = \frac{1}{e^{\frac{\omega}{T}}-1}$$
where the planck constant and the boltzmann constant $k$ is set to 1. To conservative bosonic particles, such as the core of helium-4, then it is believed to form a Bose-Einstein Condensate. I am dealing with the non-conservative bosonic particles, such as photons and phonons.
What would happen if I set the temperature $T$ equals to zero for the non-conservative bosons? My advisor believes that there would be no Bose-Einstein Condensate because the boson can pop in and out of the system. Is this true? If it is, what happens to the bosonic particles at the--or at least near--zero temperature?

Update and Edit
I've consulted my adviser and my colleagues, and this is the result. In no way that this is the solution, but it is one step further nonetheless. I've made a model of the BE distribution and it follows similarly to the image in the middle:

(source: universe-review.ca) 
What I did is vary T = 1 K, 0.1 K, and 0.01 K in Scidavis (a numerical software on linux, if you are wondering). As the T decreases, the graph gets steeper and steeper; analytically, inserting T = 0 in the formula would equal to infinity. This of course, is the puzzling question.
Since I am required to put this in my calculation, my adviser suggested that at T = 0 the distribution equals to 0, where we assumed that the photons disappear after being absorbed by the electrons.
Why does this matter so much? Because I am now calculating the self energy of the system and the final formula requires a Hilbert transform integration from -inf to +inf. If there are anyone working on this problem or something similar, this would really help.
 A: It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it does not matter.
A: The Bose-Einstein occupancy at energy $E$ is given by:
$$ f_{\mathrm{BE}}  = \frac{1}{\mathrm{e}^{\frac{E-\mu}{k_BT}}-1}.$$
The chemical potential $\mu$ is the Lagrange multiplier associated with a fixed particle number $N$. What it means is that you can vary the temperature $T$, and $\mu(T$) will also vary so as to conserve your total number of particles $N$.
Photons do not have a chemical potential, and indeed your first formula is the BE occupancy with $\mu = 0$ and $E \propto \omega$.
Thermal  photons do  not  become  quantum  degenerate: as $T\rightarrow 0$, photons vanish instead of forming a condensate. Indeed, black-body radiation ceases for an object at $T=0$.
A zero chemical potential means photons do not thermalise, which the key to achieve equilibrium and hence Bose-Einstein condensation. A laser isn't a BEC -- it's a coherent state all right, but it is not in equilibrium. The chemical potential stuff is derived in the context of equilibrium thermodynamics. So lasers are a completely different story.
In order to achieve true BEC in photons, you need to introduce photon-photon interactions, leading to photon thermalisation and a non-zero chemical potential. This has been achieved in nonlinear microcavities, dye-filled cavities and other forms of materials that interact with the photons thereby mediating the interaction amongst the photons.
