\frac{f_{m,b}(e^{-\alpha_n+i\theta})}{f_{m,b}(e^{-\alpha_n})}
 =\exp\{(e^{i\theta}-1)e^{-\alpha_n}F_{m,b}^\prime(e^{-\alpha_n})
 \nonumber \\
 & & +\frac{1}{2}(e^{i\theta}-1)^2 e^{-2\alpha_n}F_{m,b}^{\prime\prime}(e^{-\alpha_n})
 +O(\mid\theta\mid^3
 F_{m,b}^{\prime\prime\prime}(e^{-\alpha_n}))\}. \nonumber
 \end{aligned}$$ Hence, we can rewrite (\[jone\]) as follows: $$\label{jonei}
 J_1(m,n)
 =\frac{e^{n\alpha_n}}{\sqrt{2\pi}}e^{F_{m,b}(e^{-\alpha_n})}I_n,$$ where $$\begin{aligned}
 & & I_n=\frac{1}{\sqrt{2\pi}} \int_{-\delta_n}^{\delta_n}
 \exp\{(e^{i\theta}-1)e^{-\alpha_n}F_{m,b}^\prime(e^{-\alpha_n})
 \nonumber \\
 & & +\frac{1}{2}(e^{i\theta}-1)^2 e^{-2\alpha_n}F_{m,b}^{\prime\prime}(e^{-\alpha_n})
 +O(\mid\theta\mid^3
 F_{m,b}^{\prime\prime\prime}(e^{-\alpha_n}))-i\theta n\}d\theta.
 \nonumber
 \end{aligned}$$ Lemma 7 shows that, for those integers $m$ satisfying (\[mn\]), we can replace the derivatives $F_{m,b}^{(j)}(e^{-\alpha_n})$ by $F_b^{(j)}(e^{-\alpha_n}), j=1,2,3,$ at the expense of a negligible error term. In fact, combining Lemma 7 with (\[delta\]), we have $$\begin{aligned}
 & & (e^{i\theta}-1)^j F_{m,b}^{(j)}(e^{-\alpha_n}) =(e^{i\theta}-1)^j
 F_b^{(j)}(e^{-\alpha_n})
 +O(\delta_n^j\alpha_n^{-j}\log^{\rho+j}{\alpha_n^{-1}}) \nonumber
 \\
 & & =(e^{i\theta}-1)^j
 F_b^{(j)}(e^{-\alpha_n}) +O(\alpha_n^{\rho
 j/3}\log^{\rho+j}{\alpha_n^{-1}}/\omega^j(n)), \quad j=1,2,
 \nonumber
 \end{aligned}$$ and $$O(\mid\theta\mid^3 F_{m,b}^{\prime\prime\prime}(e^{-\alpha_n})) =
O(\mid\theta\mid^3 F_b^{\prime\prime\prime}(e^{-\alpha_n}))+
 O(\alpha_n^\rho(\log^{\rho+3}{\alpha_n^{-1}})/\omega^3(n)),$$ where the function $\omega(n)\to\infty$ as $n\to\infty$ arbitrarily slowly. Since all error terms above tend to $0$, we obtain $$\begin{aligned}
& & I_n =\frac{1+o(1)}{\sqrt{2\pi}} \int_{-\delta_n}^{\delta_n}
 \exp\{F_b(e^{-\alpha_n+i\theta}) -F_b(e^{-\alpha_n}) -i\theta n\}
 d\theta \nonumber \\
 & & =\frac{1+o(1)}{\sqrt{2\pi}} \int_{-\delta_n}^{\delta_n}
 \frac{f_b(e^{-\alpha_n+i\theta})}{f_b(e^{-\alpha_n})} e^{-i\theta n}d\theta, \nonumber
 \end{aligned}$$ where the last equality follows from a similar Taylor expansion for $F_b(e^{-\alpha_n+i\theta})$ and the fact that $O(\mid\theta\mid^3 F_b^{\prime\prime\prime}(e^{-\alpha_n}))
 =O(\delta_n^3 F_b^{\prime\prime\prime}(e^{-\alpha_n}))=O(1/\omega^3(n))=o(1)$ (see (\[terz\]) of Lemma 2 and (\[delta\])). Now, from Lemma 3 it follows that $$\begin{aligned}
 & & I_n \sim \frac{1}{\sqrt{2\pi}} \int_{-\delta_n}^{\delta_n}
 e^{-\theta^2 \mathcal{B}_b(e^{-\alpha_n})/2} d\theta \nonumber \\
 & & =\frac{1}{\sqrt{2\pi\mathcal{B}_b(e^{-\alpha_n})}}
 \int_{-\delta_n\sqrt{\mathcal{B}_b(e^{-\alpha_n})}}^{\delta_n\sqrt{\mathcal{B}_b(e^{-\alpha_n})}}
 e^{-y^2/2}dy \sim\frac{1}{\sqrt{2\pi\mathcal{B}_b(e^{-\alpha_n})}}
 \int_{-\infty}^\infty e^{-y^2/2}dy \nonumber \\
 & 