= Solution
{c}
Let $X_f$ have the specified <probability density function> $f_m$. Splitting the <integral> into the positive and negative half-lines, and using the <Gamma integral> $\int_0^\infty x^re^{-sx}\,dx=r!/s^{r+1}$ for a nonnegative <integer> $r$ and $s>0$, gives
$$
M_f(t)=\frac{2^m}{m!}\int_0^\infty x^m\bigl(e^{-(2-t)x}+e^{-(2+t)x}\bigr)\,dx
=2^m\left(\frac1{(2-t)^{m+1}}+\frac1{(2+t)^{m+1}}\right),\quad |t|<2.
$$
At $t=2$ or $t=-2$, one term becomes the divergent <integral> of $x^m$; beyond these endpoints it grows exponentially. Consequently, the full extended-valued <moment-generating function> is
$$
\boxed{M_f(t)=\begin{cases}2^m\bigl((2-t)^{-m-1}+(2+t)^{-m-1}\bigr),&|t|<2,\\+\infty,&|t|\geq2.\end{cases}}
$$
In particular $M_f(0)=1$, checking <normalization of a probability distribution>. All <absolute moments> exist. Direct integration yields
$$
\mathbb E|X_f|^k=\frac{2^{m+1}}{m!}\frac{(m+k)!}{2^{m+k+1}}=\frac{(m+k)!}{2^k m!}.
$$
The <symmetric distribution> makes every odd <moment> zero, so
$$
\boxed{\mathbb EX_f^k=\begin{cases}0,&k\text{ odd},\\(m+k)!/(2^k m!),&k\text{ even},\end{cases}\qquad k\geq1.}
$$
Now let $X_g$ have <probability density function> $g_m$. Its <moment-generating function> would be
$$
M_g(t)=\frac{m+1}{2}\int_1^\infty x^{-m-2}\bigl(e^{tx}+e^{-tx}\bigr)\,dx.
$$
For every $t\ne0$, one summand has exponential growth, which dominates the polynomial decay. Hence
$$
\boxed{M_g(0)=1,\qquad M_g(t)=+\infty\quad(t\ne0).}
$$
There is no finite <moment-generating function> on any neighborhood of zero, but some <moments> still exist. Specifically,
$$
\mathbb E|X_g|^k=(m+1)\int_1^\infty x^{k-m-2}\,dx
=\frac{m+1}{m+1-k}\quad\text{if }k<m+1.
$$
Thus
$$
\boxed{\mathbb EX_g^k=\begin{cases}0,&k\text{ odd and }k<m+1,\\(m+1)/(m+1-k),&k\text{ even and }k<m+1.\end{cases}}
$$
For even $k\geq m+1$, the <moment> is $+\infty$, so it is not a finite <moment>. For odd $k\geq m+1$, both the positive and negative contributions have infinite <integral>, so the <expected value> is undefined. Cancellation in a <Cauchy principal value> does not define an <expected value>. In particular, when $m=0$ there are no finite positive-order <moments>.
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