$E(X^r) = r!\{(-1)^r+(1/2)^r\}/2\;\;$ and Its PDF Let r-th moment $E(X^r) = r!\{(-1)^r+(1/2)^r\}/2\;\;$ where $r \in \Bbb N\cup\{0\} $
Then moment generating function $M(t)$ is  
$M(t) =\sum_{r=0}^\infty \dfrac{E(X^r)}{r!}t^r= \dfrac{1}{2}\left[\sum_{r=0}^\infty(-1)^rt^r+\sum_{r=0}^\infty\ ({1\over 2})^rt^r\right] = \dfrac{1}{2}\left[\dfrac{1}{1+t} + \dfrac{1}{1-{1\over 2}t}\right] = \int_0^\infty e^{tx}\cdot{1\over 2}e^x+ [?]$
I can't figure out which pdf would the last term will be.
I need to evaluate the proper pdf corresponds to above $E(X^r)$
Any hint?
 A: Consider independent but not necessarily identically distributed random variables $X_1, X_2, \ldots, X_n$, with densities $$f_{X_1}(x), f_{X_2}(x), \ldots, f_{X_n}(x),$$ and weights $a_1, a_2, \ldots, a_n \in [0,1]$ such that $\sum_{i=1}^n a_i = 1$.  Then the mixture distribution with density $$f_Y(x) = \sum_{i=1}^n a_i f_{X_i}(x)$$ is clearly a density.  The MGF of $Y$ is expressible as the weighted sum of the MGFs of the $X_i$s; i.e., $$M_Y(t) = \int_{y\in\Omega} e^{ty} f_Y(y) \, dy = \sum_{i=1}^n a_i \int_{y\in\Omega} e^{ty} f_{X_i}(y) \, dy = \sum_{i=1}^n a_i \operatorname{E}[e^{tX_i}] = \sum_{i=1}^n a_i M_{X_i}(t)$$ for the continuous case.
Since you know that $M_{X_1}(t) = (1+t)^{-1}$ corresponds to the MGF of the density $$f_{X_1}(x) = \begin{cases} e^x, & x < 0 \\ 0, & x \ge 0, \end{cases}$$ (that is to say, $-X_1 \sim \operatorname{Exponential}(\lambda = 1)$), and $M_{X_2}(t) = (1-t/2)^{-1}$ corresponds to the MGF of the density $$f_{X_2}(x) = \begin{cases} 2e^{-2x}, & x > 0 \\ 0, & x \le 0, \end{cases}$$ and the weights are simply $a_1 = a_2 = 1/2$, we find that the random variable characterized by the given MGF has the density $$f_X(x) = \begin{cases}e^{-2x}, & x > 0 \\ e^x/2, & x < 0. \end{cases}$$  The discontinuity at $x = 0$ simply means the CDF is not smooth at this point, but the CDF is unique and well-defined everywhere.
