I am trying to show that for a one dimensional Brownian motion $B_t$, ${Ee^{\zeta B_t}}$ is analytic using Morera's theorem, and $$Ee^{\zeta B_t}=e^{t \zeta^2/2}$$ for all $\zeta \in \mathbb{C}$.
The given facts are that $Ee^{i\xi B_t}=e^{-t\xi^2/2}$ for all $t\ge 0$ and $\xi \in \mathbb{R}$.
Also, using the fact that $e^{|c||y|}e^{-y^2/2} \le e^{c^2}e^{-y^2/4}$ is integrable, I get that $E e^{|\zeta|\cdot |B_t|}<\infty$ for all $\zeta \in \mathbb{C}$.
So now, since $e^{t\zeta^2/2}$ is analytic and agrees with $Ee^{\zeta B_t}$ on the real line, the proof is complete if I show that $Ee^{\zeta B_t}$ is also analytic.
I have learned that this can be proven using Morera's theorem, but that requires Fubini and I don't know how to show that the assumptions for Fubini hold. Actually I don't know how Fubini works for complex functions. How do I prove this theorem?
I have asked this question before For a one-dimensional Brownian motion $B_t$ $Ee^{\zeta B_t}=e^{t\zeta ^2/2}\; \text{for all} \; \zeta \in \mathbb{C}.$ but there is no discussion on the analyticity of $Ee^{\zeta B_t}$.