Let $X\in L^1(\Omega)$ and $\phi_X$ the corresponding characteristic function.
We know that: $\phi_X$ is $n$ times differentiable (at $u=0$) iff $\mathbb{E}[X^n]<\infty$. (This depends a bit on if $n$ is even or odd but that's not important for my question). In fact, the derivatives of $\phi_X$ give a way of computing moments, i.e. $\mathbb{E}[X^n]=i^{-n}\phi_X^{(n)}(0)$.
My questions are about avoiding to compute these derivatives and working in the complex plane instead, i.e. so-called ''generalized'' characteristic functions defined on strips of the complex plane (''strips of regulatory'').
Questions
- Suppose $X=\ln(Y)$ has a nice $C^\infty$ characteristic function. Can I compute the moments of $Y=e^X$ by simply evaluating $\phi_X$ on the imaginary axis, i.e. $$\mathbb{E}[Y^n]=\mathbb{E}[e^{i(-in)\ln(Y)}]=\phi_{\ln(Y)}(-in).$$
- If one knows that $\mathbb{E}[X^n]<\infty$ up to some $N$ (potentially infinity) due to the differentiability of $\phi_X$, does one then know that the lower half of the complex plane (up to $N$) is a subset of the stip of regularity?
Example
An example would be the Variance Gamma process (a subordinated Brownian motion). Here, \begin{align*} \phi_{X_t}(u)&=\left(\frac{1}{1-\theta\mu iu+\frac{1}{2}\mu\sigma^2u^2}\right)^{\frac{t}{\mu}} \\ &= \exp\left(-\frac{t}{\mu}\ln\left(1-\theta\mu iu+\frac{1}{2}\mu\sigma^2u^2\right)\right). \end{align*} Let $Y_t=e^{X_t}$ be an exponentiated VG process. Do we then really have \begin{align*} \mathbb{E}[Y^n] &=\phi_X(-in) \\ &=\exp\left(-\frac{t}{\mu}\ln\left(1-\theta\mu n-\frac{1}{2}\mu\sigma^2n^2\right)\right)<\infty \end{align*} for all $n\in\mathbb{N}$?