Generate Moments of Continuous Uniform Distribution with Moment Generating Functions I am having trouble generating moments from the moment generating function of the uniform.
By the definition of M.G.F, we can calculate:
$$
M(t) = \begin{cases}
\frac{e^{tb} - e^{ta}}{tb-ta} : t \ne 0 \\
1 : t=0
\end{cases}
$$
However, generating moments involves taking the nth derivative and then setting t=0. If I take the derivative of the $t \ne 0$ case, the derivative is not defined when $t=0$; if I take the derivative of the $t=0$ case, the derivative is 0. The mgf is not generating moments for uniform distribution. 
 A: As you say, the derivatives of $M(t)$ are not defined at $t=0$.
For $t\ne 0$, the first derivative for example is $$M'(t)=\frac{1}{t^2(b-a)}\left[e^{tb}(tb-1)-e^{ta}(ta-1)\right]$$
But note that $M'(t)\to \frac{a+b}{2}$ as $t\to 0$, so $M'(t)$ has a removable discontinuity at $t=0$.
So just like $M(t)$ itself, we define 
$$M'(t)=\begin{cases}
\frac{1}{t^2(b-a)}\left[e^{tb}(tb-1)-e^{ta}(ta-1)\right]&,\text{ if }t\ne 0
\\\frac{a+b}{2}&,\text{ if }t=0
\end{cases}$$
Hence the first moment is given by $$E[X]=M'(0)=\lim_{t\to 0}M'(t)$$
If you do this from definition, you will end up with the same result:
$$M'(0)=\lim_{t\to 0}\frac1t \left[M(t)-M(0)\right]=\frac{a+b}{2}$$
We define the $r$th order derivative $M^{(r)}(t)$ similarly so that it is continuous at $0$.
And the $r$th order moment of $X$ for $r\in \mathbb N$ is given by
$$E[X^r]=M^{(r)}(0)=\lim_{t\to 0}M^{(r)}(t)$$
A: The numerator is an entire function, which means you can expand it as a Taylor series around any point you like, it will converge absolutely, and you can compute with this (infinite) sum term by term.  Since for any $z,$
$$e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \cdots = \sum_{n=0}^\infty \frac{z^n}{n!},$$
compute the Taylor series of the numerator term by term.  Since the first term is always $1,$ the first term of the difference is $1-1=0,$ allowing us to begin the sum at $n=1$ instead of $n=0.$  You will quickly notice that every term in the difference is a multiple of $bt-at,$ so we may factor this out:
$$\eqalign{
e^{bt} - e^{at} &=  \sum_{n=0}^\infty \left(\frac{(bt)^n}{n!} - \frac{(at)^n}{n!}\right)\\
&=  \sum_{n=1}^\infty \frac{(bt)^n-(at)^n}{n!}  \\
&= \sum_{n=1}^\infty (bt-at) \frac{(b^{n-1}+b^{n-2}a + \cdots + b a^{n-2} + a^{n-1})t^{n-1}}{n!} \\
&= (bt-at)\sum_{n=1}^\infty  \frac{(b^{n-1}+b^{n-2}a + \cdots + b a^{n-2} + a^{n-1})t^{n-1}}{n!}.
}$$
Therefore $M$ is uniquely defined at $0$ as $M(0)=\lim_{z\to 0}M(z)$ and that limit takes no work at all to compute because 
$$M(t) = \frac{e^{bt}-e^{at}}{bt - at} =  \sum_{n=1}^\infty  \frac{(b^{n-1}+b^{n-2}a + \cdots + b a^{n-2} + a^{n-1})t^{n-1}}{n!}  =  \sum_{n=0}^\infty  \frac{(b^{n+1}-a^{n+1})}{(b-a)(n+1)}\frac{t^n}{n!}$$
also is an entire function (even in the case $b=a,$ by the way).  You can read the $n^\text{th}$ moment directly off the last expression because it is the coefficient of $t^n/n!,$ given by

$$\mu_n = \frac{b^{n+1}-a^{n+1}}{(b-a)(n+1)}.$$

Although technically we did take a limit, we did not have to compute it, and neither did we need to compute any derivatives.  (The expansion of $e^z$ is the definition of the exponential function: see Walter Rudin, Real and Complex Analysis, 1986.)

Let's check.  The first few of these moments are
$$\mu_0 = 1;\ \mu_1 = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}{2};\ \mu_2 = \frac{b^3-a^3}{3(b-a)} = \frac{b^2+ab+a^2}{3}.$$
We can easily compute these from the corresponding (raw) moments of a Uniform$(0,1)$ distribution, which are $1,$ $1/2,$ and $1/3,$ respectively, because when a variable with probability element $f(x)\mathrm{d}x$ is scaled by a factor $\sigma,$ $\mu_n$ is multiplied by $\sigma^n$ and when a variable is shifted by an amount $a$ the new moment is given by the Binomial Theorem as
$$\int_{\mathbb{R}} (x+a)^nf(x)\,\mathrm{d}x = \sum_{i=0}^n \binom{n}{i}a^{n-i}\,\int_{\mathbb{R}} x^if(x)\,\mathrm{d}x = \sum_{i=0}^n \binom{n}{i}a^{n-i}\mu_i.$$
Thus, scaling by $\sigma=b-a$ and shifting by $a$ gives
$$\eqalign{
\mu_0 &= 1\\
\mu_1 &= a(b-a)^0\mu_0 + (b-a)^1\mu_1 = a(1)+\frac{b-a}{2} = \frac{a+b}{2}\\
\mu_2 &= a^2(b-a)^0\mu_0 + 2a(b-a)^1\mu_1 + (b-a)^2\mu_2 = a^2(1) + \frac{2a(b-a)}{2} + \frac{(b-a)^2}{3} \\
&= \frac{b^2+ab+a^2}{3},
}$$
confirming the expressions given by $M.$  You can see how the calculations for higher moments are going to involve algebraic simplification of ever more complicated polynomials: the moment generating function approach spared us that work.
