I'm studying BEC and came across this result (book "Huang K Statistical Mechanics 2 edition", page 292)
Based on this expression for the internal energy:
The upper value is when v>vc and the lower one is when v<vc.
$g_{n}(z)$ are polylogarithms.
I started to derivate my internal energy, but I don't know how to proceed from: $\frac{d g_{5/2}(z)}{dz}\frac{dz}{dT}$, I can solve $\frac{d g_{5/2}(z)}{dz}$, how can I solve $\frac{dz}{dT}$?
z is obtained graphically:
For $T > T_{\text{c}}$, you are looking at the bottom expression in all three sets of equations.
So: $$ \frac{U}{N} = \frac{3}{2}\frac{k T v}{\lambda^3}g_{5/2}(z),$$ but $$ \frac{\lambda^3}{v} = g_{3/2}(z),$$ so: $$ \frac{U}{N} = \frac{3}{2}k T \frac{g_{5/2}(z)}{g_{3/2}(z)},$$
The trick is to work this out: $$ \left ( \frac{\partial z}{\partial T} \right)_V = \left [ \frac{\partial z}{\partial z g_{3/2}(z)} \frac{\partial z g_{3/2}(z)}{\partial T} \right]_V = -\frac{3 z}{2 T} \frac{g_{3/2}(z)}{g_{1/2}(z)}.$$
Because then: $$ \frac{C_v}{N k} = \frac{1}{k} \left (\frac{\partial (U/V)}{\partial T} \right )_V = \frac{1}{k} \frac{\partial}{\partial T} \left (\frac{3}{2}k T \frac{g_{5/2}(z)}{g_{3/2}(z)} \right )_V.$$
Then, with symbolic integration on Mathematica: $$ -\frac{3 T g_{\frac{1}{2}}(z(T)) g_{\frac{5}{2}}(z(T)) z'(T)}{2 z(T) g_{\frac{3}{2}}(z(T)){}^2}+\frac{3 g_{\frac{5}{2}}(z(T))}{2 g_{\frac{3}{2}}(z(T))}+\frac{3 T z'(T)}{2 z(T)},$$
and using the derivative of $z$ that I have derived earlier:
$$ \frac{C_v}{N k} = \frac{15}{4}\frac{g_{5/2}(z)}{g_{3/2}(z)} - \frac{9}{4}\frac{g_{3/2}(z)}{g_{1/2}(z)},$$
and if you plug in the expression of the root $\frac{\lambda^3}{v} = g_{3/2}(z)$ :
$$ \frac{C_v}{N k} = \frac{15}{4}\frac{v}{\lambda^3}g_{5/2}(z) - \frac{9}{4}\frac{g_{3/2}(z)}{g_{1/2}(z)}.$$
