Find Moment Generating Function of Random Variable X in which the Probability Distribution Function is:
$$f(x) = \begin{cases}1, & \text{for 0<x<1} \\
0, & \text{elsewhere}
\end{cases}$$
I understood the Moment-Generating Function to be $$M_X(t) = E[e^{(tx)}] = \int_{-\infty}^{\infty}e^{tx}f(x)dx = \int_{x=0}^{x=1} e^{tx}*1dx$$
Solving for this integral (and double-checking via Wolfram Alpha) I was able to determine that
$$\int_{x=0}^{x=1} e^{tx}*1dx \\= \frac{1}{t} e^{tx}|^1_0 \\ \quad= \frac{e^t}{t} – \frac{1}{te^0} \\= \frac{e^t-1}{t}$$
Unfortunately, the solution in the back of the book is as follows… $\text {(John E. Freund's Mathematical Statistics with Applications, 8th Edition)}$
$$M_X(t) = \frac{2e^t}{3-e^t}$$
I have no idea where this answer is coming from or where I am going wrong in terms of the Moment Generating Function equation and could definitely use some help! Thanks!
Best Answer
As noted in the comments, the work in the question is correct, except for the minor detail that the case $t=0$ should be handled separately. Since $\lim_{t\to0}\frac{e^t-1}{t}=1=M_X(0)$, this is a removable singularity.