The number of fermions in the trap is

\[
N(T, \mu)=\int \frac{d \varepsilon \varepsilon^{2}}{2(\hbar \omega)^{3}} \frac{1}{e^{\beta(\varepsilon-\mu)}-1}=\int_{0}^{\varepsilon_{F}} \frac{d \varepsilon \varepsilon^{2}}{2(\hbar \omega)^{3}}=\frac{\varepsilon_{F}^{3}}{6(\hbar \omega)^{3}} .
\]

Using \(k T_{F}=\varepsilon_{F}\) this gives the following relation for the fugacity \(z=e^{-\beta \mu}\),

\[
3\left(\frac{T}{T_{F}} \right)^{3} \int \frac{x^{2} d x}{e^{x} e^{-\beta \mu}+1}=1
\]

The internal energy is

\[
U(T, \mu)=\int \frac{d \varepsilon \varepsilon^{3}}{2(\hbar \omega)^{3}} \frac{1}{e^{\beta(\varepsilon-\mu)}-1}=\frac{(k T)^{4}}{2(\hbar \omega)^{3}} \int \frac{x^{3}}{e^{x} e^{-\beta \mu}-1} .
\]

When compared to the ground state energy \(U_{0}=\left(k T_{F} \right)^{4} /\left[8(\hbar \omega)^{3} \right]\), we get

\[
\frac{U}{U_{0}}=4\left(\frac{T}{T_{F}} \right)^{4} \int \frac{x^{3}}{e^{x} e^{-\beta \mu}-1}
\]