Asymptotics of the quotient of two integrals

For $n \in \mathbb{N}$ let $f_n : (0, \infty) \to (0, \infty)$, \begin{equation*} f_n( s) := (4y (1-y))^{2 n}, \quad y := \exp ( -\log (2) t) \end{equation*} and \begin{equation*} c_{1,n} := \int_0^\infty s^{-1} f_n (s) d s, \quad c_{2,n} := \int_0^\infty (s-1)^2 s^{-1} f_n (s) d s. \end{equation*} Proposition 1: We have \begin{equation*} \lim_{n \to \infty} (2 n)^{\frac{3}{2}} c_{2,n} = \frac{\sqrt{\pi}}{ 2\log(2)^3}. \end{equation*} Proof: The idea for the proof is to apply Lebesgue’s theorem of dominated convergence to conclude that \begin{equation*} \lim_{n \to \infty} (2 n)^{\frac{3}{2}} c_{2,n} = \int_{-\infty}^\infty t^2 \exp ( - \log(2)^2 t^2) dt = \frac{\sqrt{\pi}}{ 2\log(2)^3}. \end{equation*} Let $n \in \mathbb{N}$. Substituting with $s \mapsto \sqrt{2 n} (s-1)$ shows that \begin{equation} \label{intAfterSub} c_{2,n} = \frac{1}{(2n)^{\frac{3}{2}}} \int_{- \sqrt{2n}}^\infty \frac{t^2}{1+\frac{t}{\sqrt{2n}}} f_n \bigg(1+\frac{t}{\sqrt{2n}}\bigg) dt. \end{equation} For all $y \in \mathbb{R}$ we have \begin{equation*} 4y (1-y) = -4 \bigg(y-\frac{1}{2}\bigg)^2 +1. \end{equation*} And so for all $t \in (- \sqrt{2n}, \infty)$ it is true that \begin{equation*} \log \bigg(f_n \bigg(1+\frac{t}{\sqrt{2n}}\bigg) \bigg) = 2n \log \big( 1- h_n(t) \big), \end{equation*} where \begin{equation*} h_n (t) := \bigg(\exp \bigg( - \log (2) \frac{t}{\sqrt{2n}}\bigg)-1\bigg)^2. \end{equation*} From the Taylor expansion of the logarithm and the Lagrange form of the remainder it follows that there is $M_1 \in (0, \infty)$ such that for all $x \in [-1/2,1/2]$ it holds that \begin{equation} \label{logEstimate} |\underbrace{\log (1+x) - x}_{ =:R_1(x)} |\leq M_1 x^2. \end{equation} Similarly it follows that there is $M_2 \in (0,\infty)$ such that for all $x \in [-1,1]$ we have \begin{equation} \label{expEstimate} |\underbrace{(\exp (x) - 1)^2 - x^2}_{ =:R_2(x)} |\leq M_2 |x|^3. \end{equation} Let $t \in \mathbb{R}$ and $n \in \mathbb{N}$ such that $- \sqrt{2n} <t$. Then \begin{equation*} 2n \log \big( 1- h_n(t) \big) = 2n (- h_n(t) + R_1(-h_n(t))) \end{equation*} and \begin{equation*} - 2n h_n(t) = - 2n \log (2)^2 \frac{t^2}{2n} - 2n R_2\bigg( - \log (2) \frac{t}{\sqrt{2n}}\bigg). \end{equation*} Now combining \eqref{logEstimate} and \eqref{expEstimate} shows that \begin{equation*} \lim_{n \to \infty}R_1(-h_n(t))) =0, \ \lim_{n \to \infty}2n R_2\bigg( - \log (2) \frac{t}{\sqrt{2n}}\bigg)=0. \end{equation*} Therefore \begin{equation*} \lim_{n \to \infty} 2n \log \big( 1- h_n(t) \big) = - \log(2)^2 t^2. \end{equation*} This shows that the integrand in \eqref{intAfterSub} converges pointwise to $t \mapsto t^2 \exp ( - \log(2)^2t^2)$ for $n \to \infty$.

To apply Lebesgue’s theorem of dominated convergence it is left to find a majorant for the integrand. Recall that for all $x \in \mathbb{R}$ we have \begin{equation*} 1+x \leq e^x. \end{equation*} This implies, for all $x \in \mathbb{R}$, that \begin{equation} \label{logIneq} \log( 1+x ) \leq x \end{equation} and \begin{equation} \label{ExpSqIneq} -(e^x-1)^2 \leq -x^2. \end{equation} Let $n \in \mathbb{N}$ and $t \in ( - \sqrt{2n}, \infty)$. Then using the inequalities \eqref{logIneq} and \eqref{ExpSqIneq} shows that \begin{equation*} 2n \log \big( 1- h_n(t) \big) \leq - 2n h_n(t) \leq -2n \log(2)^2 \frac{t^2}{2n}. \end{equation*} This shows that the integrand in \eqref{intAfterSub} is majorized by $t \mapsto t^2 \exp ( - \log(2)^2 t^2)$. $\square$

Proposition 2: We have \begin{equation*} \lim_{n \to \infty} \sqrt{2 n} c_{2,n} = \frac{\sqrt{\pi}}{ \log(2)}. \end{equation*} Proof: The idea for the proof is to apply Lebesgue’s theorem of dominated convergence to conclude that \begin{equation*} \lim_{n \to \infty} \sqrt{2 n} c_{2,n} = \int_{-\infty}^\infty \exp ( - \log(2)^2 t^2) dt = \frac{\sqrt{\pi}}{\log(2)}. \end{equation*} The remainder of the proof is pretty much the same as the proof of Proposition 1. $\square$

Corollary: We have \begin{equation*} \lim_{n \to \infty} 4 n \log(2)^2 \frac{c_{2,n}}{c_{1,n}} = 1. \end{equation*} Proof: For all $n \in \mathbb{N}$ it holds that \begin{equation*} 2n\frac{c_{2,n}}{c_{1,n}} = \frac{(2n)^{\frac{3}{2}}c_{2,n}}{ \sqrt{2n} c_{1,n}}. \end{equation*} Therefore Proposition 1 and 2 and the rules for the calculation of limits show that \begin{equation*} \lim_{n \to \infty} 2n\frac{c_{2,n}}{c_{1,n}} = \frac{1}{2 \log(2)^2}. \ \square \end{equation*}