Detailed solution

a$.$ Starting from the expression for the internal energy $U(T)$ of a degenerate electron gas $($using $S=\frac{1}{2}),$

$U(T)=\frac{V}{2{}^2}(\frac{2m}{{}^2}{)}^{\frac{3}{2}}{\displaystyle \underset{0}{\overset{}{}}}\frac{E^{\frac{3}{2}}dE}{e^{(E-)}+1}$

and assuming the number of electrons is given by$,$

$N_e=\frac{V}{2{}^2}(\frac{2m}{{}^2}{)}^{\frac{3}{2}}{\displaystyle \underset{0}{\overset{}{}}}\frac{E^{\frac{1}{2}}dE}{e^{(E-)}+1}$

demonstrate that at low temperatures one can expand $U(T)$ in a series,

$U(T)=\frac{3}{5}N{E}_F[1+\frac{5{}^2}{12}(\frac{T}{T_F}{)}^2-\frac{5{}^4}{96}(\frac{T}{T_F}{)}^4+$dots$]$

Here $T_F=\frac{E_F}{k_B}.$ Obtain the terms explicitly up to order $T^2.$

To solve this problem, first integrate $U(T)$ given above by parts. Then expand $E^{\frac{5}{2}}$ in a Taylor's series about $E=.$ The integral you obtain will involve contributions dominated by energies near $.$ You should apply the same approach to the integral for $N_e,$ which will allow you to determine a relationship between $$ and $E_F.$ You will therefore need to show,

$\frac{}{E_F}$~~$1-(\frac{{}^2}{12})(\frac{T}{T_F}{)}^2+(\frac{7{}^4}{960})(\frac{T}{T_F}{)}^4+$dots

You should find this explicitly up to up to order $T^2.$ In these you do not need to find the exact results up to $T^4.$