which will be explained in detail later. The upper half plane is usually used in free probability; however we choose ${\mathbb{C}_{-}}$ to define a correspondence between a Fourier transform and a Stieltjes transform. The following observation on Cauchy distributions will be helpful to understand the reason.

It is well known that $\mathcal{F}_{\mu_{a,b}}$ and $G_{\mu_{a,b}}$ are calculated as $$\begin{aligned}
&\mathcal{F}_{\mu_{a,b}} (z)= e^{iaz - b|z|},~~~z \in \operatorname*{\mathbb{R}}, \notag \\
&G_{\mu_{a,b}}(z) = \frac{1}{z - a +ib ~\text{sign} ({\text{\normalfont Im~}}z)},  ~~~ z \in \operatorname*{\mathbb{C}}\backslash \operatorname*{\mathbb{R}},  \notag\end{aligned}$$ where $\text{sign}(x) = 1$ for $x > 0$ and $\text{sign}(x) = -1$ for $x < 0$. $\{ \mu_{at, bt} \}_{t \geq 0}$ is easily shown to become a convolution semigroup in tensor, free, Boolean and monotone senses, and therefore $\mu_{a,b}$ is infinitely divisible.

Moments of a probability measure can be obtained as the coefficients of the Taylor expansion of $\mathcal{F}_\mu (z)$ or the Laurent expansion of $G_\mu (z)$. More precisely, let $m_n(\mu)$ be the $n$th moment of $\mu$ and then $$\begin{aligned}
&\mathcal{F}_{\mu} (z)= \sum_{n=0}^\infty \frac{m_n(\mu)}{n!}(iz)^n, \notag \\
&G_{\mu}(z) = \sum_{n=0}^\infty \frac{m_n(\mu)}{z^{n+1}}.   \notag\end{aligned}$$ We neglected here the convergence of the series.

Let $\operatorname*{\mathbb{C}_{+}}$ be the upper half plane. $G_{\mu_{a,b}}$ is analytic in ${\mathbb{C}_{-}}$ and it has analytic continuation from 