Given the moment generating function of a continuous-type r.v, how to find the p.d.f?

Question

Say for $t<1$:

$$M(t) = \frac{1}{(1-t)^2}$$

How to find the p.d.f of the random variable?

$$M(t) = E(e^{tx})=\int_{-\infty}^{+\infty}e^{tx}f(x)dx$$

How do we find:

$f(x) = xe^{-x}$ on $(0,+\infty)$ and zero elsewhere? Is there a general method?

Answer 1

Replace $t$ with $it$ to get the CF, then compute the inverse Fourier transform of the CF: that gives the density. For short, the density is the inverse Laplace transform of the moment generating function. It is useful to exploit integration by parts and some well-known (inverse) transform, together with the fact that the Fourier transform of a product is a convolution of two Fourier transforms.