I do know that a power law distribution can extend from 0 to $+\infty$, so due to the shape of the distribution, there is no way to define an average value (this might be a characteristic length scale if I understood it correctly?)
But, what if a power law is truncated? still has no characteristic length?
I think we can definitely calculate $E[x] = \int_{a}^{b} xf(x) dx$, because
$f(x) = (1-\alpha)\frac{x^{-\alpha}}{b^{1-\alpha} - a^{1-\alpha}}$.
Is the $E[x]$ a characteristic length?
The problem with power law is that integral of the type $$ \int_0^{+\infty}\frac{1}{x^\alpha} dx $$ diverges in infinity, if $\alpha<1$ and diverges at zero, if $\alpha>1$. Thus, the powerl law cannot be normalized, and also has divergent moments. However, it is trancated at the limit where it diverges, it becomes normalizable, although it may still have divergent moments or other averages. If trancated at both limits, all the moments are finite, as can be easily shown by direct calculation.
