I (1) < e α 2 ∫ p − αp γ p − α √ p x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e B$_{1}$$^{x}$d x < e α $^{2}$p np 2 ( 1 − 2 B$_{1}$ n ) np 2 ∫ 1 − α $_{2}$p $^{γ}$− 1 0 y np 2 − 1 $_{e}$− np 2 $^{y}$d y, I (2) > ∫ p + αp γ p − αp γ x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e B$_{1}$$^{x}$d x > p np 2 ( 1 − 2 B$_{1}$ n ) np 2 ∫ 1+ α $_{2}$p $^{γ}$− 1 1 − α $_{2}$p $^{γ}$− 1 y np 2 − 1 $_{e}$− np 2 $^{y}$d y, and I (3) < e α 2 ∫ p + α √ p p + αp γ x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e B$_{1}$$^{x}$d x < e α $^{2}$p np 2 ( 1 − 2 B$_{1}$ n ) np 2 ∫ ∞ 1+ α $_{2}$p $^{γ}$− 1 y np 2 − 1 $_{e}$− np 2 $^{y}$d y. ∫ 1 − α $_{2}$p $^{γ}$− 1 0 y np 2 − 1 $_{e}$− np 2 $^{y}$d y = o (1) ∫ 1+ α $_{2}$p $^{γ}$− 1 1 − α $_{2}$p $^{γ}$− 1 y np 2 − 1 $_{e}$− np 2 $^{y}$d y , and ∞ 1+ α $_{2}$p $^{γ}$− 1 y np 2 − 1 $_{e}$− np 2 $^{y}$d y = o (1) 1+ α $_{2}$p $^{γ}$− 1 1 − α $_{2}$p $^{γ}$− 1 y np 2 − 1 $_{e}$− np 2 $^{y}$d y, I (0) = (1 + o (1)) ∫ p + αp γ p − αp γ x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e p ( B$_{1}$ x $_{p}$+ B$_{2}$ ( x p − 1 ) $^{2}$) d x = (1 + o (1)) p + αp γ p − αp γ x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e B$_{1}$$^{x}$d x = (1 + o (1)) ∫ p + α √ p p − α √ p x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e B$_{1}$$^{x}$d x. I (0 , ∞ ) = ∫ p + α √ p p − α √ p x np 2 − 1 $_{e}$− n $^{$_{2}$x}$e p ( B$_{0}$ + B$_{1}$ ( x p − 1 )) × ( r ∏ j =1 j ∏ s =1 √ 1 − 4 ( θ$_{pj}$v$_{pj}$ ) ( θ$_{ps}$v$_{ps}$ ) c$_{p}$ + o (1) ) d x , and hence, only constant and linear terms in the expansion of ˜ θ$_{pj}$ ˜ v$_{pj}$ − 1 2 p ∑ p i $_{=1}$ln ( 1+2 ˜ θ$_{pj}$ ˜ v$_{pj}$ − 2 ˜ θ$_{pj}$λ$_{pi}$ ) into power series of x p − 1 matter for the evaluation of I (0 , ∞ ) . Let us find these terms. This implies that Using the Laplace approximation, we have so that I (2) dominates I (1) and I (3) and Splitting the domain of integration into segments [ p − α √ p, p − αp γ ] , [ p − αp $^{γ}$, p + αp $^{γ}$] and [ p + αp $^{γ}$, p + α √ p ] , where 0 < γ < 1 / 2 , and denoting the corresponding inte- grals by I $^{(1)}$, I(2) and I $^{(3)}$, respectively, we have 36