# Moment generating function for the uniform distribution

Attempting to calculate the moment generating function for the uniform distrobution I run into ah non-convergent integral.

Building of the definition of the Moment Generating Function

$M(t) = E[ e^{tx}] = \left\{ \begin{array}{l l} \sum\limits_x e^{tx} p(x) &\text{if$X$is discrete with mass function$p( x)$}\\ \int\limits_{-\infty}^\infty e^{tx} f( x) dx &\text{if$X$is continuous with density$f( x)$} \end{array}\right.$

and the definition of the Uniform Distribution

$f( x) = \left\{ \begin{array}{l l} \frac{ 1}{ b - a} & a < x < b\\ 0 & otherwise \end{array} \right.$

I end up with a non-converging integral

$\begin{array}{l l} M( t) &= \int\limits_{-\infty}^\infty e^{tx} f(x) dx\\ &= \int\limits_{-\infty}^\infty e^{tx} \frac{ 1}{ b - a} dx\\ &= \left. e^{tx} \frac{ 1}{ t(b - a)} \right|_{-\infty}^{\infty}\\ &= \infty \end{array}$

I should find $M(t) = \frac{ e^{tb} - e^{ta}}{ t(b - a)}$, what am I missing here?

-

The density is $\frac{1}{b-a}$ on $[a,b]$ and zero elsewhere. So integrate from $a$ to $b$. Or else integrate from $-\infty$ to $\infty$, but use the correct density function. From $-\infty$ to $a$, for example, you are integrating $(0)e^{tx}$. The same is true from $b$ to $\infty$. The only non-zero contribution comes from $$\int_a^b\frac{1}{b-a}e^{tx}\,dx.$$
The limits of integration are not correct. You should integrate from $a$ to $b$ not from $-\infty$ to $+\infty$.
This answer might be correct, but it must be clear that the mistake was in writing $$\int_{-\infty}^{+\infty}e^{tx}{1\over b-a}\,dx$$ while $$\int_{-\infty}^{+\infty}e^{tx}f(x)\,dx$$ is correct. – G. Sassatelli May 18 '15 at 11:28