\end{aligned}$$ and the Laplace exponent of $\eta^+$ is $$\begin{aligned}
\label{Equation Laplace exponent phi of eta^+}
        \phi_+(q)=\kappa(\omega_++q),\quad q\geq 0.
    \end{aligned}$$ Since $\kappa'(\omega_+)>0$, the process $\eta^+$ diverges to $\infty$ almost surely, which is therefore also the case of $Y^+$ (but does not explode in finite time since $\alpha<0$).

It is possible to make sense of $Y^+$ starting from 0 as the limit of $\mathbb{P}_x^+$ as $x\to 0+$. We can thus define $\mathcal{P}_0^+$, as the law of the growth-fragmentation whose Eve cell has law $\mathbb{P}_0^+$, see Corollary 4.4 in [@BBCK16]. Note that in this latter result, the convergence of $\mathcal{P}_x^+$ towards $\mathcal{P}_0^+$ as $x\to 0$ is proved only in the sense of finite dimensional distributions. This does not allow us, for instance, to study $A$ under $\mathcal{P}_0^+$ as the limit of $A$ under $\mathcal{P}_x^+$

To emphasize with which process we are working, we shall write $\mathbb{P}_x^+$ (respectively $\mathbb{P}_x^-$) for the law of $Y^+$ (respectively $Y^-$) started from $x>0$, and $\mathbb{E}_x^+$ (respectively $\mathbb{E}_x^-$) for the induced expectation operator.

Main results {#Sectionmain results}
============

Before stating the results of this paper, we make a last assumption on the parameters. Recall that $\Lambda$ denotes the Lévy measure of $\xi$. We shall assume that its left-tail $\overline{\Lambda}:x\mapsto\Lambda({\left(-\infty\,,-x\right)})$ 