What is the pressure that supports boson star? What is the pressure that supports boson star? I noticed that for a Bose-Einstein condensate
$$
p = k_BT\frac{g}{\lambda^3}\zeta(5/2)
$$
where $g$ is the degeneracy and $\lambda$ is the thermal wavelength. The energy associated this pressure $E = \frac{3}{2}pV$ scales as $R^3$, but the gravitational energy scales as $R^{-1}$, so this pressure is not enough to counter the gravitational force. Then does it mean the star will collapse? Or is there any other contribution to the pressure?
 A: What you write down is not the pressure of a BEC, but that of a free Bose gas at zero chemical potential (you also missed a factor of m). The pressure of a non-interacting BEC is equal to this result only at $T=T_c$, where the condensate fraction is zero. Below $T_c$, the BEC is a mixture of a superfluid component (with zero pressure), and a normal component (at low $T$ we have $P\sim T^4$). Above $T_c$ the chemical potential is not zero.
Because the superfluid component has no pressure, there is nothing that can stabilize the gas in an attractive potential (either an external potential, or gravity). The gas will not usually collapse into a black hole, because during the collapse it reheats and explodes. This has been seen in cold atomic gases and is called a Bose-Nova. 
Stable BEC's require a repulsive interaction between the atoms. At the mean field level this contributes
$$
 P \sim \frac{a n^2}{m}
$$
where $a$ is the scattering length and $n$ is the density. This pressure is sufficient to stabilize atomic BEC's, and it would be sufficient to stabilize hypothetical Bose stars ($P\sim n^2$ is more repulsive than the Fermi pressure $P\sim n^{5/3}$ that stabilizes white dwarfs and neutron stars). 
A: Boson Stars are resistant to gravitational collapse due to the Heisenberg uncertainty principle. 
If you are interested, please refer to:
Stability of boson stars by Marcelo Gleiser
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.38.2376
A: The pressure of a Bose-Einstein condensate $p = kT\frac{g}{\lambda^3}\zeta(5/2)$ that composes a star, which is an interesting conjecture in a way, is countered by gravity. The hydrostatic condition for a star is
$$
\frac{dp}{dr} = \frac{GM(r)\rho(r)}{r^2}
$$
The temperature is set by the number of particles, which for a star is considerable $n \sim 10^{60}$, as
$$
T = \frac{2\pi\hbar^2}{mk}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}
$$
The only thing one has to play with are the degeneracy and the thermal wavelength. I have no particular idea what $dg/dr$ and $d\lambda/dr$ looks like. I think the degeneracy will be the crucial factor so I will propose that $g = g(r) = g_0r^\gamma$. This is then put into the $dp/dr$ on the left. The density of a BEC is another factor I can't say for certain what it is, but what dependency there is there is put into the density in the right hand side with gravity. In an iterative model one obtains $M(r)$ and works from there.
This is not entirely absurd in a way. Red dwarf stars will cool into helium gas balls and as the universe becomes very cold may in fact become BEC. This might happen for at least some part of the helium. 
