= Solution
Set $x=\nu_lT$. Expanding $e^{-x}=1-x+x^2/2-x^3/6+\cdots$ gives $1-e^{-x}=x+O(x^2)$ and $1-e^{-x}(1+x)=x^2/2+O(x^3)$. The numerator contribution is therefore $A_l\nu_lT^2/2+O(T^3)$ and the denominator contribution is $A_l\nu_lT+O(T^2)$.
For one mode, cancellation yields
$$
\boxed{\mathcal T(T)=\frac1\nu-\frac{T}{e^{\nu T}-1},\qquad\mathcal T(T)\sim T/2\ (T\to0),\qquad\mathcal T(T)\to1/\nu\ (T\to\infty).}
$$
The curve rises monotonically from zero, initially with slope $1/2$, and rounds towards a horizontal plateau. On logarithmic axes its short-time slope is one. The crossover occurs at observation times of order $1/\nu$.
Back to article page
