Power law is Nested in Power Law with Cutoff? I'm reading "Power-Law Distributions in Empirical Data": https://arxiv.org/pdf/0706.1062.pdf.
The authors make the claim that "In some cases the distributions we wish to compare
may be nested, meaning that one family of distributions is a subset of the other. The
power law and the power law with exponential cutoff in Table 2.1 provide an example
of such nested distributions."
If you were to write just the "kernels" of the pdfs for the corresponding distributions, this seems obvious:
$$ f(x) \propto x^{-\alpha}e^{-\lambda x} \ \ \text{   power law with cutoff}$$
and 
$$ g(x)\propto x^{-\alpha}  \ \ \text{power law} .$$
Sure, sure, if we let $\lambda = 0$, then $f(x)$ seems to equal $g(x)$.  But let's look at the full pdfs,
$$ f(x) = \frac{\lambda^{1-\alpha}}{\Gamma(1-\alpha, \lambda x_{min})}x^{-\alpha}e^{-\lambda x} \ \ \text{   power law with cutoff}$$
and 
$$ g(x)= (\alpha-1)x_{min}^{\alpha -1 }x^{-\alpha}  \ \ \text{power law} .$$
where $\alpha > 1$ and $x \ge x_{min} >0$.
Now it's not so obvious.  If we plug in $\lambda = 0$, then $f(x) = 0$.  
In what sense can we say that these are nested?  I am not satisfied with saying "but the normalizer doesn't matter" because, as far as the equivalence of the functions is concerned, it seems to matter a lot.
 A: As you correctly noted, the authors of this publication refer to the cases where $\alpha > 1$. With this in mind, we arrive at the following observations:


*

*When $\lambda = 0$, we divide by zero and get an indeterminate result.

*When $\lambda \rightarrow 0$, we get $f(x) \rightarrow g(x)$.
We can show this with simple asymptotic considerations: 
For $\lambda \rightarrow 0$ we get $e^{- \lambda x} \rightarrow 1$ ($x > 0$) and $\lambda ^{1-\alpha} \rightarrow \infty $ as well as $ \Gamma\left(1-\alpha,\lambda x_{min}\right) \rightarrow \infty $.
BUT: 
$$\frac{\lambda ^{1-\alpha}}{\Gamma\left(1-\alpha,\lambda x_{min}\right)} \rightarrow \frac{\alpha - 1}{x_{min}} \left(\frac{1}{x_{min}}\right)^{-\alpha} \text{, for }\lambda \rightarrow 0$$
The reason for the latter is the asymptotic behavior of the upper incomplete gamma function (https://en.wikipedia.org/wiki/Incomplete_gamma_function#Asymptotic_behavior):
$$ \frac{\Gamma\left(1-\alpha,\lambda x_{min}\right)}{(\lambda x_{min})^{1-\alpha}} \rightarrow -\frac{1}{1-\alpha} , $$ therefore
$$\frac{\lambda ^{1-\alpha}}{\Gamma\left(1-\alpha,\lambda x_{min}\right)} = \left(\frac{\Gamma\left(1-\alpha,\lambda x_{min}\right)}{(\lambda x_{min})^{1-\alpha}} (x_{min})^{1-\alpha}\right)^{-1}\rightarrow  \frac{-(1-\alpha)}{x_{min}^{1-\alpha}} =\frac{\alpha - 1}{x_{min}} \left(\frac{1}{x_{min}}\right)^{-\alpha}. $$
In this sense, power law and power law with exponential cutoff are nested distributions as you can derive the former from the latter for $\lambda \rightarrow 0$.
In addition, the log-log plot below shows how the power law with cutoff asymptotically approaches the power law in the arbitrary case of $x = 1$, $\alpha = 2.5$, $x_{min} = 0.1$ and $\lambda = 10^{-m}$.

