Ideal Bose condensation gas in gravitational field, Statistical Physic

[dpopchev]

Homework Statement

We have an ideal Bose gas in gravitational field. Show that the critical temperature( the temperature in which the condensation is starting is:
$T_c = T^0_c( 1 + \frac{8}{9} \frac{1}{ζ(3/2)} (\frac{\pi mgh}{kT^0_c})^{1/2} )$

Attempt
Lege artis; Firstly I want to estimate the number $N$ of particles in that $T_c$ temperature. As we know at that state the chemical potential is zero so $\mu = 0$, so the average number of particles will be given by the Bose - Einstein distribution
$n_{mean} = \frac{1}{e^{ε/kT_c}-1}$
I want to find all that particles in such state, so I form an integral over the volume in which the gas is( we assume it is a container with axial symmetry, with area of the circle $S$ and hight $L$ ) and over all possible energy states. I will skip the mathematical formulation and give the integral I need to evaluate:
$N = 4 \pi g_0 S h^{-3} \int_0^{∞} dp \int_0^L dz \frac{p^2}{e^{ \frac{p^2}{2 m k T_c} - \frac{mgz}{kT_c} } - 1 }$
where $S$ is area of the circle for the cylinder, $g_0 = 2s + 1$ is the number of states and $h$ is the Planck constant
After changing the variables I get $\frac{ 4 \pi g_0 S (2mkT_c)^{3/2}kT_c}{h^3mg} \int^{∞}_0 dp' \int^{\frac{mgL}{kT_c}}_0 \frac{1}{e^{p'^2}e^{-z'}-1}$
I have no idea how to handle this integral:
I just get the $e^{z'}$ out and get $\int \frac{e^{z'}}{e^{p'^2}-e^{z'}}$ which leads to something like $\int p'^2 ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } dp'$ on which my best attempt was $\int ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } \frac{dp'^3}{3} = \frac{p'^3}{3}.ln(...) - ...$ and the first is in range of $0$ which is a problematic dot and $∞$ which doesn't seem pretty too.
I hope for my post to be accurate enough. Thanks in advance.

PS: I found the same problem but it can't help me to solve the integral
https://www.physicsforums.com/showthread.php?t=455803

 

[Asim Riaz]

The Attempt at a SolutionWe have an ideal Bose gas in gravitational field. Show that the critical temperature( the temperature in which the condensation is starting is: T_c = T^0_c( 1 + \frac{8}{9} \frac{1}{ζ(3/2)} (\frac{\pi mgh}{kT^0_c})^{1/2} ) Solution: We can use the Bose-Einstein distribution to calculate the average number of particles, N, at the critical temperature, T_c. The Bose-Einstein distribution states that the average number of particles at temperature T is given by N = \frac{1}{e^{\frac{\epsilon}{kT}} - 1}where \epsilon is the energy of the particle. We can then use this to calculate the total number of particles in the system by integrating over the volume and energy states. The integral is given by N = 4 \pi g_0 S h^{-3} \int_0^{∞} dp \int_0^L dz \frac{p^2}{e^{\frac{p^2}{2m k T_c} - \frac{mgz}{kT_c}} - 1} where g_0 = 2s + 1 is the number of states, S is the area of the cylinder, and h is the Planck constant. By changing variables we can rewrite the integral as N = \frac{4 \pi g_0 S (2m k T_c)^{3/2}k T_c}{h^3 mg} \int_{0}^{\infty} dp' \int_{0}^{\frac{mgL}{kT_c}} \frac{1}{e^{p'^2}e^{-z'} - 1} We can then use the fact that \int_{0}^{\infty} \frac{e^{z'}}{e^{p'^2}-e^{z'}} dp' = \frac{\pi}{\sqrt{2}} \frac{