I read this post How is the minimum of a set of IID random variables distributed? where I can find how to compute the density distribution of the minimum between $N$ positive random variables. If the cumulative of these random variables is $F(x)$ the cumulative of the minimum between $N$ of these random variables is:
$$ \mathbb{P}(\text{min}\le x)=1-\big(1-F(x)\big)^N$$
From this result it is easy to compute the mean value of the minimum (simply using integration by parts): $$ \overline{\text{min}}=\int_{0}^{\infty}x\,\frac{d \mathbb{P}(\text{min}\le x)}{dx}dx=\int_{0}^{\infty}\big(1-F(x)\big)^Ndx $$ At this point I can compute $\overline{\text{min}}$ for a general cumulative distribution $F(x)$.
I want now to get the scaling of $\overline{\text{min}}$ for $N\rightarrow\infty$ when $F(x)=\frac {1}{r!}\int_{0}^{x}l^re^{-l}dl$ for $r\in\mathbb{Z}$ and $r>-1$. (A simple case can be the one for $r=0$, that is the density distribution for the variables is exponential, in this case the average min value scale as $1/N$). In the general case I can arrive to: $$ \overline{\text{min}}=\int_{0}^{\infty}\left(1-\frac{1}{r!}\int_0^{x}l^re^{-l}dl\right)^Ndx $$ but I don't know how to approximate it. For general $r$ it can be useful to approximate $\frac{l^re^{-l}}{r!}\simeq \frac{l^r}{r!}\propto l^r$, since the important contribution for the average minimum value will be $\ll1$. In this approximation my result is: $$ \overline{\text{min}}=\int_{0}^{\infty}\left(1-\int_0^{x}l^rdl\right)^Ndx $$ but still can't figure out which is the scaling with large $N$.
The result should be: $$ \overline{\text{min}}\simeq N^{-\frac{1}{r+1}} $$
I am studying the following paper: https://hal.science/jpa-00232897/document .
Here the authors find the scaling of the mean value of what I suppose is the minimum of the random variables, which in this case are $l_{ij}$, called “distances”. I cannot find the scaling found after Eq.(5), can anyone help me?
