g(x) = \int_0^\infty f(x/t)dg(t), \quad x>0. \end{aligned}$$ we denote the *Mellin-Stieltjes convolution* for functions $f$ and $g$ such that the Stieltjes integral is well defined.

\[funkcja-miara\] Let $\alpha\in(0,2)$. ${\operatorname{Re}}\psi $ is regularly varying at infinity (at $0$) with the exponent $\alpha$ if and only if $t\mapsto {\nu}(\{s \colon|s|{\geqslant}t\})$ is regularly varying at 0 (at infinity) with the exponent $-\alpha$. Moreover, $$\begin{aligned}
\nu(\{s\colon|s|{\geqslant}t\})  \cong  \frac{\Gamma(1 + \alpha) }{\mathrm{B}\Big(1 - \frac{\alpha}{2},1 + \frac{\alpha}{2}\Big)} {\operatorname{Re}}\psi(1/t), \quad t \to 0^+ \quad (t\to \infty).\end{aligned}$$

We compute the Laplace transform of the function ${\operatorname{Re}}\psi$. By Fubini’s theorem, $$\begin{aligned}
\mathcal{L}\{{\operatorname{Re}}{\psi}\}(\lambda) = & \int_0^\infty e^{-\lambda t} \int_ {\mathbb{R}}(1-\cos tx)\,{\nu}(dx)\,dt = \int_{\mathbb{R}}\int_0^\infty e^{-\lambda t}(1- \cos tx)\,dt\,{\nu}(dx) \\
                      = & \frac{1}{\lambda}  \int_{\mathbb{R}}\frac{x^2}{\lambda^2 + x^2}\,{\nu}(dx).\end{aligned}$$ Next, $$\begin{aligned}
\frac{\lambda}{2} \mathcal{L}\{{\operatorname{Re}}\psi\} (\lambda) = &\frac{1}{2} \bigg( \int_{-\infty}^{0}\frac{x^2}{\lambda^2 + x^2}\,\nu(dx) +  \int_{0}^{\infty} \frac{x^2}{\lambda^2 + x^2}\,{\nu}(dx) \bigg) \\ = & -\int_{-\infty}^0   \int_x^0 \frac{\lambda^2 t}{ (\lambda^2 +t^2)^2 }\,dt \,\nu(dx) + \int_0^\infty   \int_0^x \frac{\lambda^2 t}{ (\lambda^2 +t^2)^2 }\,dt\,\nu(dx) \\ = & -\int_{-\infty}^0 \frac{ (\lambda t)^2}{ (\lambda^2 +t^2)^2 } \nu((-\infty,t])  \frac{dt}{t} + \int_0^\infty \frac{(\lambda t)^2}{ (\lambda^2 +t^2)^2 }  \nu([t,\infty))  \frac{dt}{t} \\ = & \int_0^\infty \frac{  (\frac{\lambda}{t})^2 }{{ (1 + (\frac{\lambda}{t})^2)}^2 } \big( {\nu}((-\infty,-t]) +{\nu}([t,\infty)) \big)\frac{dt}{t}.\end{aligned}$$

Assume that ${\operatorname{Re}}\psi$ varies regularly at infinity with exponent $\alpha\in (0,2)$. Let us define 