In this paper they use an inversion formula for characteristic functions whose argument is complex with non-zero real part. Namely, given a random variable $X$ taking values in $(0,\infty)$, they consider the characteristic function in the following way \begin{align*} f(\phi)=\mathbb{E}\left[e^{\phi X}\right]=\mathbb{E}\left[e^{\phi_R X+ i\phi_I X}\right], \end{align*} where $\phi=\phi_R+i\phi_I\in\mathbb{C}$ and $\phi_R,\phi_I\in\mathbb{R}$. Note that $f$ may not be well-defined, as opposed to the standard characteristic function that is always well-defined. Then, in Equation (A8) of the paper they obtain the density function $p$ of the random variable $X$ using the following inversion theorem \begin{align*} p(x)=\frac{1}{\pi} \int_0^\infty\Re\left[e^{-\phi x}f(\phi)\right]d\phi_I. \end{align*} I would like to know if there is some reference for this result or if it can be deduced from the existing inversion theorems for standard characteristic functions. For instance, the previous inversion formula looks like a generalization of the Gil-Pelaez inversion formula, that is, if we consider the characteristic function $\varphi(t)=\mathbb{E}\left[e^{itX}\right]$ where now $t\in\mathbb{R}$, then, \begin{align*} p(x)=\frac{1}{\pi} \int_0^\infty\Re\left[e^{-it x}\varphi(t)\right]dt. \end{align*}
So far, I have been able to prove that \begin{align*} \frac{1}{\pi} \int_0^\infty\Re\left[e^{-\phi x}f(\phi)\right]d\phi_I=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\phi_Rx-i\phi_Ix}\mathbb{E}\left[e^{\phi_RX+i\phi_IX}\right]d\phi_I \end{align*} Note that if $\phi_R=0$ in the previous expression we recover the usual inversion formula given by \begin{align*} p(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-i\phi_Ix}\mathbb{E}\left[e^{i\phi_IX}\right]d\phi_I \end{align*}
Any help on how to prove this inversion theorem or any reference of the inversion theorem used in the paper would be really appreciated.
Finally, I found a way to prove it using existing inversion theorems. I think this should work:
Let $X$ be a real-valued random variable and $\phi_R\in\mathbb{R}$ such that $\mathbb{E}\left[e^{\phi_R X}\right]<\infty$. Write $f(\phi)=\mathbb{E}\left[e^{\phi X}\right]$ where $\phi=\phi_R+i\phi_I\in\mathbb{C}, \phi_R,\phi_I\in\mathbb{R}$ is the generalized Fourier transform. We assume that $t\to f(\phi_R+it)$ is integrable in $[0,\infty)$ and that $X$ has a density function $p$.
Define the random variable $Y$ that has density function given by \begin{align*} p_Y(y)=\frac{1}{K}e^{\phi_R y}p(y) \end{align*} where $K=\int_{-\infty}^\infty e^{\phi_R x}p(x)dx=\mathbb{E}\left[e^{\phi_R X}\right]<\infty$. Note that the characteristic function of $Y$ is given by \begin{align*} f_Y(i\phi_I)&=\mathbb{E}\left[e^{i\phi_I Y}\right]=\int_{-\infty}^\infty e^{i\phi_I y}p_Y(y)dy=\frac{1}{K}\int_{-\infty}^\infty e^{i\phi_I y}e^{\phi_R y}p(y)dy \end{align*} \begin{align*} &=\frac{1}{K}\mathbb{E}\left[e^{\left(\phi_R+i\phi_I\right) X}\right]=\frac{1}{K}f(\phi_R+i\phi_I). \end{align*} Since $t\to f(\phi_R+it)=f_Y(it)$ is integrable in $[0,\infty)$, by the Gil-Pelaez inversion theorem the following holds \begin{align*} p_Y(y)=\frac{1}{\pi}\int_{0}^\infty \Re\left[e^{-i\phi_I y}f_Y(i\phi_I)\right]d\phi_I . \end{align*} Introducing the equivalent expressions for $p_Y$ and $f_Y$ we get \begin{align*} \frac{1}{K}e^{\phi_R y}p(y)&=\frac{1}{K\pi}\int_{0}^\infty \Re\left[e^{-i\phi_I y}f(\phi_R+i\phi_I)\right]d\phi_I \\ \end{align*} \begin{align*} p(y)&=\frac{1}{\pi}\int_{0}^\infty \Re\left[e^{-\left(\phi_R+i\phi_I\right)y}f(\phi_R+i\phi_I)\right]d\phi_I=\frac{1}{\pi}\int_{0}^\infty \Re\left[e^{-\phi y}f(\phi)\right]d\phi_I. \end{align*} This finishes the proof.