The energy of the different rotational levels are given by the rigid rotor model: $\epsilon_{rot,J}=\frac{\hbar^2J(J+1)}{2I}$, the degeneracy of the levels is $g_J=2J+1$

$$q_{rot}=\sum_{J=0}^{\infty}g_Je^{\epsilon_{rot,J}/kT}=\sum_{J=0}^ {\infty}(2J+1)e^{-\frac{\hbar^2 J(J+1)}{2IkT}}$$

We call rotational temperature $\theta_{rot}$ a: $\ theta_{rot}=\frac{\hbar^2}{2Ik}$, has units of temperature (K) but is not temperature in the physical sense. Therefore:

$$q_{rot}=\sum_{J=0}^{\infty}(2J+1)e^{-\frac{\theta_{rot}}{{T}}J( J+1)}$$

If $\frac{\theta_{rot}}{T}$ is small, the separation between rotational levels is small compared to $kT$ and we can approximate the sum by an integral.

$$q_{rot}=\int_{0}^{\infty}(2J+1)e^{-\frac{\theta_{rot}}{T}J(J+1)}dJ$$

Making the change of variable $w=J(J+1)\Rightarrow dw=(2J+1)dJ \Rightarrow dJ=\frac{dw}{2J+1}$

$$q_{ rot}=\int_{0}^{\infty}e^{-\frac{\theta_{rot}}{T}w}dw=\left[-\frac{T}{\theta_{rot}}e ^{-\frac{\theta_{rot}}{T}w}\right]_{0}^{\infty}=\frac{T}{\theta_{rot}}$$

The function of Rotational partitioning must be modified by introducing the symmetry number $\sigma$ which takes the value 1 for homonuclear molecules and 2 for heteronuclear ones.

$$q_{rot}=\frac{T}{\sigma\theta_{rot}}$$

For $T>\theta_{rot}$ the above equation makes a significant error and we must use the following development:

$$q_{rot}=\frac{T}{\sigma\theta_{rot}}\left[1+\frac{1}{3}\frac{\theta_{rot}}{ T}+\frac{1}{15}\left(\frac{\theta_{rot}}{T}\right)^2+\frac{1}{315}\left(\frac{\theta_{rot }}{T}\right)^3+....\right]$$

For $T<\theta_{rot}$ the partition function must be evaluated by summation: $q_{rot}=\sum_{J=0}^{\infty}(2J+1)e^{-\frac{\theta_{rot}}{T}J(J+1)}$