Using density of states $\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}}{{\mathbf{h\omega }}_{\mathbf{0}}}$, which generalizes equation (9-27) to account for multiple allowed spin states (see Exercise 52), the definition ${\mathbf{Nh\omega }}_{\mathbf{0}}\mathbf{/}\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{\epsilon }$and $\mathbf{N}\mathbf{=}{\int }_0^{\infty }N\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{dE}$. Solve for $\mathbf{B}$in distributions (9-32) and (9-33) careful use of $\mathbf{\pm }$will cut your work by about half. Then plug back in and show that for a system of simple harmonic oscillators, the distributions become $\mathbf{\gamma }{\mathbf{\left(}\mathbf{E}\mathbf{\right)}}_{\mathbf{\text{BE}}}\mathbf{=}\frac{\mathbf{1}}{\frac{{\mathbf{e}}^{\mathbf{E}\mathbf{/}{\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}}}{\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{\delta }\mathbf{/}{\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}}}\mathbf{-}\mathbf{1}}\mathbf{\text{and}}\mathcal{N}{\mathbf{\left(}\mathbf{E}\mathbf{\right)}}_{\mathbf{\text{FD}}}\mathbf{=}\frac{\mathbf{1}}{\frac{{\mathbf{e}}^{\mathbf{E}\mathbf{/}{\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}}}{{\mathbf{e}}^{\mathbf{+}\mathbf{\delta }\mathbf{/}{\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{1}}$.

You will need the following integral:${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{(}{\mathbf{Be}}^{\mathbf{2}}\mathbf{\pm }\mathbf{1}\mathbf{)}}^{\mathbf{-}\mathbf{1}}\mathbf{dz}\mathbf{=}$$\mathbf{\pm }\mathbf{ln}\left(1\pm \frac{1}{B}\right)$.

The Bose-Einstein statistics:

$\mathcal{N}\left(\mathrm{E}\right)=\frac{1}{{\mathrm{Be}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1}$

The Fermi-Dirac statistics:

$\mathcal{N}\left(\mathrm{E}\right)=\frac{1}{{\mathrm{Be}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}+1}.$

The normalization condition writes:

$\mathrm{N}={\int }_0^{\mathrm{\infty }}\mathcal{N}\left(\mathrm{E}\right)\mathrm{D}\left(\mathrm{E}\right)\mathrm{dE}$

Where, dE is the energy range, and $\mathrm{D}\left(\mathrm{E}\right)$is the density of states which in this problem, takes the form:

$\mathrm{D}\left(\mathrm{E}\right)=\frac{2\mathrm{s}+1}{\mathrm{\hslash }{\mathrm{\omega }}_0}$

Here,$\mathcal{N}\left(\mathrm{E}\right)$ is the occupation number, and depending on the type of particle, follows the form:

$\mathcal{N}\left(\mathrm{E}\right)=\frac{1}{{\mathrm{Be}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\pm 1},$

Let $\mathrm{z}=\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$, then we have$\mathrm{dz}=\frac{\mathrm{dE}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\Rightarrow \mathrm{dE}={\mathrm{k}}_{\mathrm{B}}\mathrm{Tdz}.$

Performing a change of variables, the integrand becomes a function of$\mathrm{z}$:

$\begin{array}{c}\mathrm{N}=\frac{2\mathrm{s}+1}{\mathrm{\hslash }{\mathrm{\omega }}_0}{\int }_0^{\mathrm{\infty }}{({\mathrm{Be}}^{\mathrm{z}}\pm 1)}^{-1}\left({\mathrm{k}}_{\mathrm{B}}\mathrm{Tdz}\right)\\ ={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{\hslash }{\mathrm{\omega }}_0}{\int }_0^{\mathrm{\infty }}{({\mathrm{Be}}^{\mathrm{z}}\pm 1)}^{-1}\mathrm{dz}\end{array}$

Notice that ${\int }_0^{\mathrm{\infty }}({\mathrm{Be}}^{\mathrm{z}}\pm 1)\mathrm{dz}$$\mathrm{is}\mathrm{a}\mathrm{familiar}\mathrm{integral}\mathrm{with}\mathrm{an}\mathrm{analytical}\mathrm{value}\mathrm{of}:$

${\int }_0^{\mathrm{\infty }}({\mathrm{Be}}^{\mathrm{z}}\pm 1)\mathrm{dz}=\pm \ln (1\pm 1/\mathrm{B})$

$\begin{array}{c}\mathrm{N}={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{\hslash }{\mathrm{\omega }}_0}(\pm \ln (1\pm 1/\mathrm{B}\left)\right)\\ =\pm {\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{\hslash }{\mathrm{\omega }}_0}\ln (1\pm 1/\mathrm{B})\end{array}$

Solving for $B$, we thus obtain its expression depending on which type of particle it is. For fermions, ${\mathrm{B}}_{\text{FD}}$is thus:

$\begin{array}{c}\mathrm{N}={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{\hslash }{\mathrm{\omega }}_0}\ln (1+1/{\mathrm{B}}_{\text{FD}})\\ \ln (1+1/{\mathrm{B}}_{\text{FD}})=\frac{{\mathrm{Nh\omega }}_0}{2\mathrm{s}+1}\frac{1}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\end{array}$

Setting$\epsilon =\mathrm{N}\mathrm{\hslash }{\mathrm{\omega }}_0/(2\mathrm{s}+1)$we get:

$\begin{array}{c}\ln (1+1/{\mathrm{B}}_{\text{FD}})=\frac{\mathcal{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ {\mathrm{e}}^{\ln (1+1/{\mathrm{B}}_{\text{FD}})}={\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ 1+1/{\mathrm{B}}_{\text{FD}}={\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ \frac{1}{{\mathrm{B}}_{\text{FD}}}={\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1\\ {\mathrm{B}}_{\text{FD}}=\frac{1}{{\mathrm{e}}^{\mathcal{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1}\cdot \end{array}$

For bosons,${\mathbf{B}}_{\text{BE}}$ is thus:

$\begin{array}{c}\mathrm{N}=-{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\frac{(2\mathrm{s}+1)}{\mathrm{\hslash }{\mathrm{\omega }}_0}\ln (1-1/{\mathrm{B}}_{\text{BE}})\\ \ln (1-1/{\mathrm{B}}_{\text{BE}})=-\frac{\mathrm{N}\mathrm{\hslash }{\mathrm{\omega }}_0}{2\mathrm{s}+1}\frac{1}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\end{array}$

Setting $\mathcal{E}=\mathrm{N}\mathrm{\hslash }{\mathrm{\omega }}_0/(2\mathrm{s}+1)$we get:

$\begin{array}{c}\ln (1-1/{\mathrm{B}}_{\text{FD}})=-\frac{\mathrm{\epsilon }}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ {\mathrm{e}}^{\ln (1-1/{\mathrm{B}}_{\text{BE}})}={\mathrm{e}}^{-\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ 1-1/{\mathrm{B}}_{\text{BE}}={\mathrm{e}}^{-\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ \frac{1}{{\mathrm{B}}_{\text{BE}}}=1-{\mathrm{e}}^{-\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ {\mathrm{B}}_{\text{BE}}=\frac{1}{1-{\mathrm{e}}^{-\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\end{array}$

Plugging the expression for${\mathrm{B}}_{\text{FD}}$into eq. (3), the occupation number associated with a system of fermionic oscillators having intrinsic spin is thus:

$\mathcal{N}{\left(\mathrm{E}\right)}_{\text{FD}}=\frac{1}{\frac{{\mathrm{e}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}{{\mathrm{e}}^{\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1}+1}$

Finally, plugging the expression for ${\mathrm{B}}_{\text{BE}}$into eq. (3), the occupation number associated with a system of bosonic oscillators having intrinsic spin is thus:

$\mathcal{N}{\left(\mathrm{E}\right)}_{\text{BE}}=\frac{1}{\frac{{\mathrm{e}}^{\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}{1-{\mathrm{e}}^{-\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}-1}$

$\begin{array}{c}{\mathrm{B}}_{\text{FD}}=\frac{1}{{\mathrm{e}}^{\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1}\\ {\mathrm{B}}_{\text{BE}}=\frac{1}{1-{\mathrm{e}}^{-\mathrm{\epsilon }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\end{array}$