I have an equation as $\mathbb{E}[e^{tXab}]$ which I know is a moment generating function (mgf) where $a$ and $b$ are just two constant numbers.
If I assume that the probability distribution is uniform on [-1,1]. Then the mgf can be easily calculated as
$M_X(t)=\mathbb{E}[e^{tX}] = \frac{e^{t}-e^{-t}}{2t} $
But I wonder how I can calculate the first equation ($\mathbb{E}[e^{tXab}]$)?
Please let me know if the problem is not clear enough.
If $\mathsf E(e^{tX})=M_X(t)$ then $\mathsf E(e^{tXab})=M_X(tab)$.
Now $M_X(t)=\tfrac{1}{2t}(e^t-e^{-t})$, so $M_X(tab)=\ldots$