$\nu_1=\nu|_{|x|{\leqslant}1}$ and the characteristic exponent $\psi_1$ corresponding to the triplet $(0,0,\nu_1)$. We have ${\operatorname{Re}}\psi\cong{\operatorname{Re}}\psi_1$ at infinity. Indeed, since $$0{\leqslant}\int_{|x|>1}(1-\cos(xz))\,\nu(dx){\leqslant}2 \nu\big(\{x\colon |x|>1\}\big),\quad z\in \mathbb{R},$$ and $\lim_{z\to\infty} {\operatorname{Re}}\psi(z)= \infty$, we get $$\frac{{\operatorname{Re}}\psi(z)-{\operatorname{Re}}\psi_1(z)}{{\operatorname{Re}}\psi(z)} \to 0,$$ as $z \to \infty$. Next, using the Abel theorem [@RegularVariation89 Theorem 1.7.1] one can observe that $$\begin{aligned}
\lambda^{-1}\mathcal{L}\{{\operatorname{Re}}{\psi}_1\}(1/\lambda) \cong {\operatorname{Re}}\psi(\lambda) \Gamma(1 + \alpha), \quad \lambda \to \infty.\end{aligned}$$

Let $g(t)= \nu_1(\{s\colon |s|{\geqslant}1/t\}),\, t > 0$ and $k(t) = \frac{t^2}{{(1+t^2 )}^2}$. Observe that the Laplace transform of ${\operatorname{Re}}\psi_1$ is the Mellin convolution of $k$ and $g$: $$\begin{aligned}
\frac{1}{2\lambda} \mathcal{L}\{{\operatorname{Re}}\psi_1\} (1/\lambda) = \int_0^\infty k(1/(t\lambda)){g}(1/t)\frac{dt}{t} =\int_0^\infty k(t/\lambda){g}(t)\frac{dt}{t}= k \stackrel{m}{*} {g}(\lambda),\end{aligned}$$ where in the last equality we used $k(t)=k(1/t)$ for $t>0$.

To prove that $g(t)$ is regularly varying function, we will use [@RegularVariation89 Theorem 4.9.1] for the function $g_1(t) = \int_t^\infty {g}(s)\frac{ds}{s}$ and convolution $ k \stackrel{s}{*} g_1(\lambda)= k \stackrel{m}{*} g(\lambda)$. Set $\sigma$ such that $ -2 <\sigma <-\alpha$ and $\tau = 0$. Observe that $$\begin{aligned}
\|k\|_{\sigma, \tau} = & \sum_{-\infty < n <\infty} \max(e^{-\sigma n}, e^{-\tau n}) \sup_{e^n {\leqslant}x {\leqslant}e^{n+1}} \big| k(x)\big| \\
{\leqslant}&\ \ \sum_{n {\leqslant}-1}  e^{2n} +  \sum_{n {\geqslant}0}  \frac{e^{-\sigma n}}{e^{2n}} < \infty.\end{aligned}$$ Moreover, by [@MR0352890 Table 1.2 (2.19)], $$\mathcal{M}\{k\}(z)=\int^\infty_0\frac{t^2}{(1+t^2)^2}t^{z-1}\,dt=\frac{1}{2}\int^\infty_0\frac{1}{(1+s)^2}s^{(z/2+1)-1}\,ds= \frac{\Gamma\big( 1-\frac{z}{2} \big)\Gamma\big( 1 + \frac{z}{2} \big)}{2}.$$ The function $\Gamma$ 