I am an undergraduate, who has just begun getting into probability theory. Recently we learned of the moment generating function of a random variable can be used to find moments, and the moment generating function is defined as
$$ MGFx(t) = E[e^{tX}] =\begin{cases} \sum e^{tx}p(x), & \text{if X is discrete}\\ \int e^{tx}f(x)dx,\ & \text{if X is continuous} \end{cases}$$ The moment generating function can also be viewed as a Laplace transform of the probability density function of the random variable X, by replacing s with -t,
$${\scr L}(pdf(X))=M_X(t)=\int_0^\infty f(x)e^{xt}\,dt$$
From here, one may take derivations with respect to t of $M_x(t)$, and then evaluate the derivative at t = 0, to find the respective moment of the probability distribution function, i.e.
$$ \frac{d^{n}}{dt^{n}}\ M_x(t) = M^n_X(t) $$ $$ M^n_X(0) = E[X^n], n =1,2,3,...$$ When I have dealt with the Laplace transform of a function in the past, for differential equations, the text stated that the transform redefines a function $f(t)$ such that the function is now defined by a complex variable s, where $s = \sigma +i\omega$, and
$${\scr L}(f(t))=\int_0^\infty f(t)e^{-st}\,dt =F(s)$$
For the majority of differential equations that I have dealt with, that involve a Laplace transform to solve, we are converting to from a function of time(t) to one of frequency(s), and the transformed function now has computation done on it in the complex frequency domain until the transform is reversed.
My question is that for the Laplace transform of a probability distribution function, what does the complex domain that the function is transformed to represent? More specifically what is the variable "t" that is introduced, and what is the meaning of the derivative of the moment generating function, with respect to t? Is the variable t arbitrary or is it the representation of something?
Thank you.