By carrying out the integration suggested just before equation (9-28), show that the average energy of a one-dimensional oscillator in the limit ${\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}\mathbf{\gg }{\mathbf{h\omega }}_{\mathbf{0}}$is${\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}$.

The expression for the average energy $\overline{\mathbf{E}}$of a system of one-dimensional harmonic oscillators is found to be:

$\overline{\mathbf{E}}\mathbf{=}\frac{{\displaystyle \int {\mathrm{ENAe}}^{-E/k_{B}T}(1/\hslash {\omega }_0)\mathrm{dE}}}{{\displaystyle \int {\mathrm{NAe}}^{-E/k_{B}T}(1/\hslash {\omega }_0)\mathrm{dE}}}$

Simplify the expression further yields:

$\overline{\mathrm{E}}=\frac{{\displaystyle \int {\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}}}{{\displaystyle \int {\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}}}$

In the limit where ${\mathrm{k}}_{\mathrm{B}}\mathrm{T}>>\mathrm{\hslash }{\mathrm{\omega }}_0$, the integration is carried out from 0 to $\infty$.

Thus:$\overline{\mathrm{E}}=\frac{{\displaystyle {\int }_0^{\mathrm{\infty }}{\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}}}{{\displaystyle {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}}}$

First evaluate the numerator. Consider:

${\int }_0^{\mathrm{\infty }}{\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}$

Use integration by parts, perform the change of variables:

Let:$\mathrm{u}=\mathrm{E};\mathrm{dv}=\int {\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}$

Then solve further:

$\mathrm{du}=\mathrm{dE};\mathrm{v}=-{\mathrm{k}}_{\mathrm{B}}{\mathrm{Te}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Use integration by parts, recall that:

$\int \mathrm{udv}=\mathrm{uv}-\int \mathrm{v}\mathrm{du}$

If:

$\int \mathrm{udv}=\int \mathrm{E}{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}$

then:

$\begin{array}{c}\int {\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}=-{\mathrm{k}}_{\mathrm{B}}{\mathrm{TEe}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-\int -{\mathrm{k}}_{\mathrm{B}}{\mathrm{Te}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}\\ =-{\mathrm{k}}_{\mathrm{B}}{\mathrm{TEe}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}+{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\int {\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}\end{array}$

Solve the integral:

$\int {\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}=-{\mathrm{k}}_{\mathrm{B}}{\mathrm{TEe}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-{\left({\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right)}^2{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Let us now evaluate the definite integral:

$\begin{array}{c}{\int }_0^{\mathrm{\infty }}{\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}=-{{\mathrm{k}}_{\mathrm{B}}{\mathrm{TEe}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}|}_0^{\mathrm{\infty }}-{{\left({\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right)}^2{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}|}_0^{\mathrm{\infty }}\\ =0-{\left({\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right)}^2(-1)\\ ={\left({\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right)}^2\end{array}$

Now consider the denominator. Consider the definite integral:

${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}$

Evaluate the problem:

$\begin{array}{c}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}={(-{\mathrm{k}}_{\mathrm{B}}\mathrm{T}){\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}|}_0^{\mathrm{\infty }}\\ =-{\mathrm{k}}_{\mathrm{B}}\mathrm{T}(-1)\\ ={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\end{array}$

Take their quotient, the average energy for a system of harmonic oscillators is thus:

$\begin{array}{c}\overline{\mathrm{E}}=\frac{{\displaystyle {\int }_0^{\mathrm{\infty }}{\mathrm{Ee}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}}}{{\displaystyle {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{E}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dE}}}\\ =\frac{{\left({\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right)}^2}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ ={\mathrm{k}}_{\mathrm{B}}\mathrm{T}\end{array}$