\lambda(\theta)&=1+\int_0^\infty (e^{i\theta x}-1)\,dG(x)+
\sum_{s=1}^4 q_s\int_0^\infty (e^{i\theta x}-1)\,dG_\theta^s(x) \\
&=1+i\theta\int_0^\infty e^{i\theta x}(1-G(x))\,dx+
\sum_{s=1}^4 q_si\theta\int_0^\infty e^{i\theta x}g_\theta^s(x)(1-G(x))\,dx,\end{aligned}$$ where $q_1=1$, $q_2=-1$, $q_3=i$, $q_4=-i$, $g_\theta^s(x)=\int_x^\infty v_\theta^s(u)\,dG(u)/
\int_x^\infty \,dG(u)\ll\theta^\beta$.

Now, $$\begin{aligned}
& i\theta \int_0^\infty e^{i\theta x}(1-G(x))\,dx 
 = 
ic\theta\int_0^\infty e^{i\theta x}x^{-\beta}\,dx +
ic\theta\int_0^\infty e^{i\theta x}H_1(x)\,dx 
\\ & \qquad = ic\theta^\beta \int_0^\infty e^{i\sigma}\sigma^{-\beta}\,d\sigma +
ic\theta\int_0^\infty H_1(x)\,dx +
ic\theta\int_0^\infty (e^{i\theta x}-1)H_1(x)\,dx \\
& \qquad = -cc_\beta\theta^\beta +icc_H \theta + ic\theta A\end{aligned}$$ where $$\begin{aligned}
A & = 
\int_0^\infty  (e^{i\theta x}-1)H_1(x)\,dx =
\int_0^{1/\theta} (e^{i\theta x}-1)H_1(x)\,dx +
\int_{1/\theta}^\infty (e^{i\theta x}-1)H_1(x)\,dx \\
& \ll \theta\int_0^{1/\theta} xH_1(x)\,dx +
\int_{1/\theta}^\infty H_1(x)\,dx 
\ll \theta\int_0^{1/\theta} x^{1-q'}\,dx +
\int_{1/\theta}^\infty x^{-q'}\,dx 
 \ll \theta^{q'-1}.\end{aligned}$$

It remains to estimate the terms $\int_0^\infty e^{i\theta x}g_\theta^s(x)(1-G(x))\,dx$. We give the details for $\int_0^\infty \sin\theta x\,g_\theta^s(x)(1-G(x))\,dx$; the case with $\sin$ replaced by $\cos$ is identical. Since $x\mapsto g_\theta^s(x)(1-G(x))$ is decreasing for each fixed $\theta$, we can write $$\begin{aligned}
 \int_0^\infty \sin\theta x\,g_\theta^s(x)(1-G(x))\,dx  
& \le
\int_0^{\pi/\theta} \sin\theta x\,g_\theta^s(x)(1-G(x))\,dx 
\\ & \ll \theta^\beta\int_0^{\pi/\theta} x^{-\beta}\,dx \ll \theta^{2\beta-1},\end{aligned}$$ giving the required upper bound, and the lower bound is obtained in the same way. 

Identification of the Fourier coefficients {#sec-Fourier}
==========================================

Let $\widehat R_n$ and $\widehat T_n$ denote the Fourier coefficients of $R(\theta)$ and $T(\theta)$. By (H1), $R$ is uniformly absolutely summable on $S^1$. Therefore $\widehat R_n=R_n$. In this section, 