"infinite moments" or "moments don't exist"? what is the difference between "infinite moment" and "moments don't exist"?
moreover, if I find out the moment generating function for some distribution, and take first derivative set s=1, and if I get $\infty$, what does it mean? infinite mean? or mean doesn't exist?
Thanks in advance.
 A: When the sum or integral defining $E[|X^n|]$ diverges, $E[X^n]$ might diverge to $+\infty$, so you could say "the $n$'th moment is $+\infty$" as an alternative to "the $n$'th moment does not exist".  This would have to be the case if $n$ is even or if $X \ge 0$ almost surely.  However, if $n$ is odd and $X$ can be either positive or negative, both positive and negative parts might diverge, so all you could say is "the $n$'th moment does not exist". 
A: If $X$ has a Cauchy distribution, then $E(X^2)=\infty$, and one sometimes expresses that by saying the second moment does not exist.  But concerning $E(X^3)$, one may say that it does not exist, but one cannot say that it is infinite.  If you look at
$$
E(X^3) = \int_{-\infty}^\infty x^3 \frac{dx}{\pi(1+x^2)},
$$
what you find is that both the positive and negative parts are infinite:
$$
\int_{-\infty}^0 x^3 \frac{dx}{\pi(1+x^2)}=-\infty\text{ and }\int_0^\infty x^3 \frac{dx} {\pi(1+x^2)} = +\infty.
$$
When that happens, you can't say the integral has any particular value, because you can actually change the value by rearraning the way in which the bounds approach infinity.  For example, if you find
$$
\lim_{a\to\infty} \int_{-a}^a \frac{x}{1+x^2}\;dx \text{ and }\lim_{a\to\infty} \int_{-a}^{2a} \frac{x}{1+x^2}\;dx,
$$
then you actually get two different finite numbers.  (That can happen only if the positive and negative parts are both infinite.)
Later note: Things like the central limit theorem and the law of large numbers fail in cases where the integral that defines the first moment has both the positive and negative parts infinite.  For example, suppose
$$
X_1,\ldots,X_N\sim\operatorname{i.i.d.}\operatorname{Cauchy},
$$
so the probability density of each is $\displaystyle f(x) = \frac{1}{\pi(1+x^2)}$.  Then the probablity distribution of the sample mean $\bar X=(X_1+\cdots+X_n)/n$ is the same Cauchy distribution.   So $\bar X$ won't converge to a point as $n\to\infty$, nor will $\sqrt{n}\;\bar X$ converge weakly to a normal distribution.
