moment-generating function of the chi-square distribution How do we find the moment-generating function of the chi-square distribution?  I really couldn't figure it out.  The integral is 
$$E[e^{tX}]=\frac{1}{2^{r/2}\Gamma(r/2)}\int_0^\infty x^{(r-2)/2}e^{-x/2}e^{tx}dx.$$
I'm going over it for a while but can't seem to find the solution.
By the way, the answer should be $$(1-2t)^{(-r/2)}.$$

In case you have not yet figure it out, the value of the integral follows by simple scaling of the integrand. First, assume $t < \frac{1}{2}$, then change variables $x = (1-2 t) y$:
$$
\int_0^\infty x^{(r-2)/2} \mathrm{e}^{-x/2}\mathrm{e}^{t x}\mathrm{d}x =
 \int_0^\infty x^{r/2} \mathrm{e}^{-\frac{(1-2 t) x}{2}} \, \frac{\mathrm{d}x}{x} =
 \left(1-2 t\right)^{-r/2} \int_0^\infty y^{r/2} \mathrm{e}^{-\frac{t}{2}}  \, \frac{\mathrm{d}y}{y}
$$
The integral in $y$ gives the normalization constant, and value of m.g.f. follows.
