the corresponding exponential functional .

Starting from , multiplying on both sides with ${\mathbb{L}}_\mu(u)=e^{-\psi_\mu(u)}$ and dividing once by $u$ we obtain for $u >0$ $$\begin{aligned}
\label{eq:densityh1}
 \frac{\psi_\eta (u)}{u}{\mathbb{L}}_\mu(u) = & -(\gamma_0 - \frac{\sigma_\xi^2}{2}) {\mathbb{L}}'_\mu(u) + \frac{\sigma^2_\xi}{2} u {\mathbb{L}}''_\mu(u)  + \int_{{\mathbb{R}}} \left( \frac{{\mathbb{L}}_\mu(ue^{-y})}{u} - \frac{{\mathbb{L}}_\mu(u)}{u} \right) \nu_\xi(dy). \end{aligned}$$ Now assume that $\mu$ has a density, such that ${\mathbb{L}}_\mu(u)=\int_0^\infty e^{-ut} f(t) dt$. Denote the inverse Laplace transform by $\overset{{\mathbb{L}}^{-1}}\longrightarrow $, then obviously we have ${\mathbb{L}}_\mu(u)\overset{{\mathbb{L}}^{-1}}\longrightarrow  f(t)$ $\lambda$-a.e. while (assuming $\lim_{t\to 0} t^2 f(t)=0$ and that $f$ is differentiable) $\lambda$-a.e. we get $$\begin{aligned}
  {\mathbb{L}}'_\mu(u) & \overset{{\mathbb{L}}^{-1}}\longrightarrow -tf(t),\\
u {\mathbb{L}}''_\mu(u)& \overset{{\mathbb{L}}^{-1}}\longrightarrow \frac{d}{dt}(t^2 f(t))= 2tf(t)+t^2 f'(t),\\
 \int_{{\mathbb{R}}} \left( \frac{{\mathbb{L}}_\mu(ue^{-y})}{u} - \frac{{\mathbb{L}}_\mu(u)}{u} \right) \nu_\xi(dy) & \overset{{\mathbb{L}}^{-1}}\longrightarrow  \int_t^\infty \nu_\xi((\log\frac{s}{t}, \infty)) f(s) ds - \int_0^t \nu_\xi((-\infty, \log\frac{s}{t}))  f(s) ds,\end{aligned}$$ where the last line follows from $$\begin{aligned}
 \lefteqn{\int_{{\mathbb{R}}} \left( \frac{{\mathbb{L}}_\mu(ue^{-y})}{u} - \frac{{\mathbb{L}}_\mu(u)}{u} \right) \nu_\xi(dy) }\\ 
&= \int_{{\mathbb{R}}} \left(\int_0^\infty e^{-ut} \left( \int_0^{t e^y} f(s) ds - \int_0^t f(s) ds  \right) dt \right) \nu_\xi(dy) \\
&= \int_0^\infty e^{-ut} \left(\int_t^\infty f(s) \left(\int_{\log\frac{s}{t}}^\infty   \nu_\xi(dy)\right) ds \right)dt - \int_0^\infty e^{-ut} \left(\int_0^t f(s) \left(\int_{-\infty}^{\log\frac{s}{t}} \nu_\xi(dy)\right) ds \right)dt.\end{aligned}$$ Further for the left hand side of with $\psi_\eta (u)=a_\eta u + \int_{(0,\infty)} (1-e^{-ut})\nu_\eta(dt)$ we will use that $$\frac{\int_{(0,\infty)} (1-e^{-ut})\nu_\eta(dt)}{u}{\mathbb{L}}_\mu(u)= \int_0^\infty e^{-us} \nu_\eta((s, \infty)) ds \, {\mathbb{L}}_\mu(u)\overset{{\mathbb{L}}^{-1}}\longrightarrow  \int_0^t 