Moment generating function of standard normal

Question

As far as I know, given a random variable $X$, we define its moment generating function as $$M_X(t) = \mathbb{E} \left[ e^{ tX} \right] \ \ , \ t \in \mathbb{R}$$ I read that MGF for a general random variable may not be defined for negative values of $t$. What about if $X \sim N(0,1)$? Can we be sure its MGF is well defined for negative t?

Answer 1

For $X \sim N(0,1)$ we have $Ee^{tX}=e^{t^{2}/2}$ for all real numbers $t$. One way to prove this is to use the characteristic function and use a basic result from Complex Analysis [The Identity Theorem].