Does any moment generating function implies an existence of moments? For a real-valued r.v. $\xi$ we suppose and existence of its moment generating function $m(t) = \mathsf E\mathrm e^{t\xi}$ for all $t\in (-h,h)$ where $h>0$. I wonder how many moments $M_n = \mathsf E\xi^n$ are finite. 
I know that using the differentiation under the Lebesgue integral, local existence of $m(t)$ implies an existence of $M_1 = \mathsf E\xi$. On the other hand for the 2nd moment it seems that the finiteness of $M_2$ is equivalent to the statement that $m''(0)$ exists, hence there should be examples when $m(t)$ in the neighborhood of $0$ while $M_2 = \infty$. 
For an example when $M_2$ does exist without an existence of $m(t)$ for $t>0$ we can consider a r.v. with a density
$$
f(x) = \frac{2}{\pi(1+x^2)^2}.
$$
Could you provide such an example when $m(t)$ exists in the neighborhood of $t=0$ but $M_2 = \infty$? Maybe you can also refer me to the literature since this question is not covered in my book on the probability.
 A: As is well known, if the moment-generating function (mgf) exists in some open interval containing $0$, then all moments are finite. Indeed, suppose that $\xi$ has a finite mgf in some open interval containing $0$. Then, there exists a $t \neq 0$ such that
$$
\int_{( - \infty ,0)} {e^{ (- |t|)x} F(dx)}  < \infty 
$$
and
$$
\int_{[0,\infty)} {e^{|t|x} F(dx)}  < \infty ,
$$
where $F$ is the distribution of $\xi$.
Given $n \in \mathbb{N}$, this implies that
$$
\int_{( - \infty ,0)} {|x|^n F(dx)}  < \infty 
$$
and 
$$
\int_{[0 ,\infty)} {x^n F(dx)}  < \infty ,
$$
respectively (note that $\int_{[ - M,M]} {|x|^n F(dx)}  < \infty $, for any $M > 0$ fixed). Hence
$$
\int_{( - \infty ,\infty)} {|x|^n F(dx)}  < \infty,
$$
so the $n$th moment of $\xi$ is finite.
EDIT (two well-known facts): 
1) The converse is not true: the lognormal distribution has finite moments of all orders, but $m(t)=\infty$ for any $t>0$.
2) In case $\xi$ has a finite mgf $m(\cdot)$ in some open interval containing $0$, then it holds
$$
m^{(n)}(0) = {\rm E}(\xi^n),
$$
and $m(\cdot)$ may be expanded in a power series about $0$ as
$$
m(t) = \sum\limits_{n = 0}^\infty  {\frac{{m^{(n)} (0)}}{{n!}}t^n }  = \sum\limits_{n = 0}^\infty  {\frac{{{\rm E}(\xi ^n )}}{{n!}}t^n } .
$$
