with the proof of free infinite divisibility of fGIG distributions.

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[**Case 1.**]{} $\lambda<1$.

Observe that we have $$\label{hypo}
{\text{\normalfont Im}}(r_{\alpha,\beta,\lambda}(x)) \leq 0, \qquad x\in{\mathbb{R}}\setminus\{\alpha\}.$$ From we see that ${\text{\normalfont Im}}(r_{\alpha,\beta,\lambda}(x))=0$ for $x \in(-\infty,\alpha) \cup (\alpha,\eta]$, and $$\label{r}
{\text{\normalfont Im}}(r_{\alpha,\beta,\lambda}(x)) = \frac{(x-\delta)\sqrt{\beta (x-\eta)}}{x (\alpha-x)}<0,\qquad x>\eta$$ since $\eta>\alpha>0>\delta$.\
Moreover observe that by for ${\varepsilon}>0$ small enough we have ${\text{\normalfont Im}}(r_{\alpha,\beta,\lambda}(\alpha+{\varepsilon}e^{i \theta}))<0$, for $\theta\in[-\pi,0]$. Now Lemma \[lem:32\] implies that free GIG distribution if freely ID in the case $\lambda>0$.

[**Case 2**]{} $\lambda>0$.

In this case similar argument shows that $\mu(\alpha,\beta,\lambda)$ is FID. Moreover by $\eqref{alpha2}$ point $z=\alpha$ is a removable singularity and $r_{\alpha,\beta,\lambda}$ extends to a continuous function on ${\mathbb{C}}^-\cup{\mathbb{R}}$. Thus one does not need to take care of the behaviour around $z=\alpha$.

[**Case 3**]{} $\lambda=0$.

For $\lambda=0$ one can adopt a similar argumentation using . It also follows from the fact that free GIG family $\mu(\alpha,\beta,\lambda)$ is weakly continuous with respect to $\lambda$. Since free infinite divisibility is preserved by weak limits, then the case $\lambda=0$ may be deduced from the previous two cases.

Next we will determine the free Lévy measure of free GIG distribution $\mu(\alpha,\beta,\lambda)$.

Let $(\xi_{\alpha,\beta,\lambda}, \zeta_{\alpha,\beta,\lambda},\tau_{\alpha,\beta,\lambda})$ be the free characteristic triplet of the free GIG distribution 