does not have any roots, so the Wiener condition $\mathcal{M}\{k\}(z) \neq 0$ is satisfied. Notice that $g_1$ is non increasing on $(0, \infty)$ and the function $g_1(t)$ is zero on $(0,1)$. Hence, $g_1(t) = \mathcal{O}(t^\sigma)$ at $0^+$. That is, the kernel $k$ and the function $g_1$ satisfy assumptions of [@RegularVariation89 Theorem 4.9.1], hence, $$\begin{aligned}
g_1(t) \cong \frac{\Gamma(1+\alpha)}{\mathrm{B}\Big(1 - \frac{\alpha}{2},1 + \frac{\alpha}{2}\Big)} \frac{{\operatorname{Re}}\psi(t)}{\alpha}, \quad t \to 0^+.\end{aligned}$$ By monotone density theorem [@RegularVariation89 Theorem 1.7.2] and the fact that $g(1/t)\cong \nu(\{s\colon |s|{\geqslant}t\})$ as $t$ goes to $0^+$, we obtain that $$\begin{aligned}
\nu(\{s\colon |s|{\geqslant}t\})  \cong  \frac{\Gamma(1 + \alpha) }{\mathrm{B}\Big(1 - \frac{\alpha}{2},1 + \frac{\alpha}{2}\Big)} {\operatorname{Re}}\psi(1/t), \quad t \to 0^+.\end{aligned}$$ In particular, $t\mapsto \nu(\{s\colon |s|{\geqslant}t\})$ is regularly varying function at $0$ with index $-\alpha$.

Now assume that $t\mapsto \nu(\{s \colon |s|{\geqslant}t\})$ is regularly varying function at $0$ with index $-\alpha$. Again, instead of $\psi$ one can consider $\psi_1$. Since $${\operatorname{Re}}\psi_1(z)= z\int^1_0 \sin(xz) \nu(\{s\colon |s|{\geqslant}x\})\,dx=\int^z_0 \sin(xz) \nu(\{s\colon |s|{\geqslant}x/z\})\,dx,$$ one can use Potter’s Lemma to justify that $$\lim_{z\to\infty}\frac{{\operatorname{Re}}\psi_1(z)}{\nu(\{s\colon|s|{\geqslant}1/z\})}=\int^\infty_0\frac{\sin x}{x^\alpha}\,dx,$$ which finishes the proof in this case.

If ${\operatorname{Re}}\psi$ varies regularly at 0 one can modify the above prove to obtain the behaviour of the tail of $\nu$ at infinity.

We remark that in fact, equivalence of regular 