Characterizing the integral as a function of $n$

Question

Let $\alpha \in [0,3], \beta \geq 1, \lambda \geq 1$ and fix $n \in \mathbb{N}$. Consider the function $f(x;\alpha, \beta, \lambda) = x^{\alpha}\exp(-\lambda x^\beta)$. Let $I(n; \alpha,\beta,\lambda) = \int_0^\infty 1 - (1 - f(x;\alpha, \beta, \lambda))^n dx$.

I want to know how does function $I(n;\alpha, \beta, \lambda)$ scale with $n$ and the parameters $\alpha, \beta, \lambda$. I know that for $\alpha = 0, \beta = 1$ and $\lambda > 0$, we have that $I(n;\alpha = 0, \beta = 1, \lambda) \sim \log n / \lambda$. This is because we have that \begin{align} I(n;\alpha = 0, \beta = 1, \lambda) &= \int_{0}^\infty (1 - (1 - \exp(-\lambda x))^n) dx \\ &= (1 / \lambda) \int_{0}^1 (1 - (1 - t)^n )/ t dt \\ &= (1 /\lambda) \sum_{k = 1}^n {n \choose k} (-1)^{k + 1} / k \\ &= (1 / \lambda) \sum_{k = 1}^n 1 / k \\ &\sim \log n / \lambda \end{align} where the last equality follows from an identity for the harmonic series I found here.

Intuitively, I conjecture that the more general case is \begin{align} I(n;\alpha, \beta = 1, \lambda) \sim (\log n / \lambda) + o((\log n)) \end{align}

However, due to the extra $x^\alpha$ in $f(x; \alpha, \beta, \lambda)$, it is unclear to me if there is a clean and simple proof for this or whether the conjecture is true in the first place?

Moreover, can one show a more general result for $\beta \geq 1$, where \begin{align} I(n;\alpha, \beta, \lambda) \sim C(\alpha, \beta, \lambda) (\log n)^{1 / \beta} - o((\log n)^{1/\beta}) \end{align}

Any help or pointers would be greatly appreciated.

Answer 1

$\newcommand{\be}{\beta}\newcommand{\al}{\alpha}\newcommand{\la}{\lambda}$The integral in question is \begin{equation*} I=I_0+I_1+I_2, \tag{10}\label{10} \end{equation*} where \begin{equation*} I_2:=\int_0^\infty dx\,1(f(x)\ge c_1)(1-(1-f(x))^n), \end{equation*} \begin{equation*} I_1:=\int_0^\infty dx\,1(c_{-1}\le f(x)<c_1)(1-(1-f(x))^n), \end{equation*} \begin{equation*} I_0:=\int_0^\infty dx\,1(f(x)<c_{-1})(1-(1-f(x))^n), \end{equation*} \begin{equation*} c_p:=\frac{\ln^p n}n,\quad f(x):=x^\al e^{\la x^\be}, \end{equation*} \begin{equation*} \al\in[0,3],\quad\be\ge1,\quad\la\ge1. \end{equation*} Everywhere here, it is assumed that $n\ge2$, and the limits are taken as $n\to\infty$.

Note that for real $x>0$ and real $p$ \begin{equation*} f(x)\ge c_p\iff x_{p,1}\le x\le x_{p,2}, \end{equation*} where $x_{p,1}$ and $x_{p,2}$ are the positive roots of the equation $f(x)=c_p$ such that $x_{p,1}\to0$, $c_p=f(x_{p,1})\sim x_{p,1}^\al$, so that \begin{equation*} x_{p,1}\sim c_p^{1/\al}=\Big(\frac{\ln^p n}n\Big)^{1/\al}; \tag{20}\label{20} \end{equation*} $x_{p,2}\to\infty$, $c_p=f(x_{p,2})=e^{-(1+o(1))\la x_{p,2}^\be}$, so that \begin{equation*} x_{p,2}\sim \Big(\frac{\ln n}\la\Big)^{1/\be}. \tag{30}\label{30} \end{equation*}

For real $x>0$ and $r:=\al/\be$, we have $0<r\le3$ and hence \begin{equation*} 0<f(x)\le\max_{u>0}u^r e^{-u}=(r/e)^r\le(3/e)^3=1.344\ldots<2. \end{equation*} So, if $x>0$, $f(x)\ge c_1$, and $n$ is large enough, then $|1-f(x)|\le1-c_1$ and hence \begin{equation*} I_2\sim\int_{x_{1,1}}^{x_{1,2}} dx\sim x_{1,2}\sim \Big(\frac{\ln n}\la\Big)^{1/\be}. \tag{40}\label{40} \end{equation*}

If $x>0$ and $f(x)<c_{-1}$, then $nf(x)\to0$ and hence $$1-(1-f(x))^n\sim nf(x),$$ so that \begin{equation*} I_0\sim n(I_{01}+I_{02}), \tag{50}\label{50} \end{equation*} where \begin{equation*} I_{01}:=\int_0^{x_{-1,1}} dx\,f(x)\sim\int_0^{x_{-1,1}} dx\,x^\al=\frac{x_{-1,1}^{\al+1}}{\al+1} \\ \sim\frac1{\al+1}\,\Big(\frac{\ln^{-1} n}n\Big)^{1+1/\al}=o(1/n), \tag{60}\label{60} \end{equation*} \begin{equation*} I_{02}:=\int_{x_{-1,2}}^\infty dx\,f(x) =\int_{x_{-1,2}^\be}^\infty\frac{du}\be\,u^{(\al+1)/\be-1} e^{-\la u} \\ =O(x_{-1,2}^{\al+1-\be} e^{-\la x_2^\be}) =O(c_{-1}x_{-1,2}^{1-\be})=o\Big(\frac{(\ln n)^{1/\be}}n\Big), \tag{70}\label{70} \end{equation*} \begin{equation*} I_1\le \int_0^\infty dx\,1(c_{-1}\le f(x)<c_1) \\ =x_{1,1}-x_{-1,1}+x_{-1,2}-x_{1,2}=o((\ln n)^{1/\be}) \tag{80}\label{80} \end{equation*} by \eqref{20} and \eqref{30}.

Collecting \eqref{10}, \eqref{40}, \eqref{50}, \eqref{60}, \eqref{70}, and \eqref{80}, we conclude that \begin{equation*} I\sim\Big(\frac{\ln n}\la\Big)^{1/\be}. \end{equation*}