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Set ${\theta }_{\mathrm{rot}}=hc\tilde{B}/k_{\mathrm{B}}$, the rotational temperature. For a continuous nonnegative variable $J$,Equating this derivative to zero gives the most populated rotational levelThe second derivative is $-4/(2J+1{)}^2-2{\theta }_{\mathrm{rot}}/T<0$, so this is a maximum when $J_{\ast }\ge 0$. At lower temperatures the constrained continuous maximum is $J=0$. For the physical integer maximum compare the nearby allowed levels explicitly.

For the supplied linear rigid rotor spectrum, sum over every nonnegative integer $J$ and include all $2J+1$ magnetic substates:It is convenient to introduce the rotational temperatureso that the first terms are $Z_{\mathrm{rot}}=1+3e^{-2{\theta }_r/T}+5e^{-6{\theta }_r/T}+\cdots$. Here $I$ is the moment of inertia and $\hslash$ is the reduced Planck constant, as confirmed in the original PDF. The stipulated degeneracies are used as given, without adding nuclear-spin weights or omitting selected $J$ values.

For a linear rigid rotor, rotational temperature expresses the rotational energy constant in kelvins: $E_J=k_B{\theta }_{r}J(J+1)$, where $I$ is the perpendicular moment of inertia. With all integer $J\ge 0$ allowed and degeneracies $2J+1$, its rotational canonical partition function is $Z={\sum }_J(2J+1)e^{-{\theta }_{r}J(J+1)/T}$. It approaches $1+3e^{-2{\theta }_r/T}$ at low temperature and $T/{\theta }_r$ at high temperature. The latter limit gives mean rotational energy $k_{B}T$ and heat capacity $k_B$ per rotor.