small. We have the Potter’s bounds [@Potter42], [@BinghamGoldieTeugels Theorem 1.5.6]: $C^{-1}\sigma^\beta\le h(\sigma/\theta)\ell(1/\theta)^{-1}\le C\sigma^{-1}$ uniformly in $\theta>0$, $\sigma\in(0,2\pi]$, where $C$ is a constant. Hence $$\begin{aligned}
A_{(i)} & \ge
  \frac{\sqrt 2}{2}\int_0^{\pi/4}e^{-\sigma y}\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1} \,d\sigma+O\Bigl(
\int_{\pi/4}^{3\pi/2}e^{-\sigma y} \,d\sigma\Bigr) 
 \\ & = \frac{\sqrt 2}{2}\int_0^{\pi/4}e^{-\sigma y}\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1} \,d\sigma+O(y^{-1}e^{-(\pi/4) y}),\end{aligned}$$ $$\begin{aligned}
A_{(ii)} & \le 2|g_z|_\infty\int_0^{\pi/4}e^{-\sigma y}\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1} \,d\sigma +O\Bigl(\int_{\pi/4}^{\pi}e^{-\sigma y} \,d\sigma\Bigr) 
\\ & \ll |g_z|_\infty A_{(i)} + O(y^{-1}e^{-(\pi/4) y}).\end{aligned}$$ Since $|g_z|_\infty=o(1)$, we can choose $\epsilon$ sufficiently small that $$|1-\lambda(z)|\gg u\ell(1/\theta)\theta^{\beta-1}
\Big\{\int_0^{\pi/4}e^{-\sigma y}\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1} \,d\sigma +O(y^{-1}e^{-(\pi/4) y})\Bigr\}.$$ Furthermore, $$\int_0^{\pi/4}e^{-\sigma y}\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1} \,d\sigma \gg \int_0^{\pi/4}e^{-\sigma y}\,d\sigma=y^{-1}+O(y^{-1}e^{-(\pi/4) y}).$$ For $y\ge 1/\delta$ with $\delta$ sufficiently small, the terms $O(y^{-1}e^{-(\pi/4) y})$ are negligible so that $1-\lambda(z)\gg u\ell(1/\theta)\theta^{\beta-1}y^{-1}=\ell(1/\theta)\theta^\beta$.

It remains to consider the complementary region $y\in(0,1/\delta]$. Note that $\int_0^{3\pi/2} e^{-\sigma y}\cos\sigma\, \sigma^{-\beta}\,d\sigma$ depends continuously on $y$ and is positive for all $y\ge0$. It follows by compactness that $\int_0^{3\pi/2} e^{-\sigma y} \cos\sigma\, \sigma^{-\beta}\,d\sigma$ is bounded away from zero for $y\in[0,1/\delta]$. Moreover there exists $b\in(0,3\pi/2)$ such that $\int_b^{3\pi/2} e^{-\sigma y} \cos\sigma\, \sigma^{-\beta}\,d\sigma$ is bounded away from zero for $y\in[0,1/\delta]$. By uniform convergence of slowly varying functions [@BinghamGoldieTeugels Theorem 1.2.1], we can shrink $\epsilon$ if necessary so that $|h(\sigma/\theta)/\ell(1/\theta)-1|$ is as small as desired, uniformly in $\sigma\in[b,3\pi/2]$ and $\theta\in(0,\epsilon]$. Hence $A_{(i)}\ge \int_0^{3\pi/2} \cos\sigma\, e^{-\sigma y}\sigma^{-\beta}h(\sigma/\theta)\ell(1/\theta)^{-1}\,d\sigma$ which is bounded away from 