How does $\langle 1/t\rangle $ denote a rate?

Question

I am going through an old paper of mathematical physics, and they claim $\langle 1/t\rangle$, where $\langle \cdot\rangle$ on an interval $[a,b]$ denotes the statistical average: $$\langle f\rangle = \frac{1}{b-a}\int_a^b f$$ is physically a rate. I can't see how the integral of the inverse of a time variable is a rate. Using the above definition, again on an interval $[a,b]$ we obtain $\langle 1/t\rangle = \log(b/a)/(b-a)$ which looks unlike any "rate" I have ever seen.