of the form .\
Similarly, the Laplace transform of a random variable $X$ on ${\mathbb{R}}_+$ with $\mu={\mathcal{L}}(X)$ is written as ${\mathbb{L}}_X(u)={\mathbb{L}}_\mu(u)={\mathbb{E}}[e^{-uX}]= e^{-\psi_X(u)}=e^{-\psi_\mu(u)}$. Please notice, that this notation of Laplace exponents is different from the previous papers [@BLexpfunc; @BLMranges] but coincides with the notation used in [@rene-book].

As in [@BLexpfunc; @BLMranges], given a one-dimensional Lévy process $(\xi_t)_{t\geq 0}$ drifting to $+\infty$, we will consider the mapping $$\begin{aligned}
\Phi_\xi^+ : D_\xi^+  & \to {\mathcal{P}}^+ ,\\
{\mathcal{L}}(\eta_1) & \mapsto {\mathcal{L}}\left( \int_0^\infty e^{-\xi_{s-}} \,
d\eta_s \right),\end{aligned}$$ defined on $$\begin{aligned}
D_\xi^+ := \{ {\mathcal{L}}(\eta_1) : \eta=(\eta_t)_{t\ge 0} \, & \mbox{ one-dimensional subordinator independent of }\xi\\
& \hskip 15mm \mbox{ such that }\int_0^\infty
e^{-\xi_{s-}} \, d\eta_s\mbox{ converges a.s.}\},\end{aligned}$$ and we denote the range of $\Phi_\xi^+$ by $$R_\xi^+ := \Phi_\xi^+(D_\xi^+).$$

On the density of the exponential functional {#sec:density}
============================================

As already observed in previous articles, it follows directly from [@alsmeyeretal Thm. 1.3] that the exponential functional $V$ has a pure-type law, i.e. its distribution is either absolutely continuous, continuous singular or a Dirac measure, where the latter can only be obtained if both processes, $\xi$ and $\eta$, are deterministic (c.f. [@BLexpfunc Prop. 6.1]).\
Absolute continuity of exponential functionals has been studied in detail in [@bertoinlindnermaller08]. For the 