Kth moment of Weibull distribution The Weibull distribution is given by parameters $c,\beta>0$ and so that for all $t\geq 0$, $P(X>t)=\exp(-ct^\beta)$. 
I want to find $E[X^k]$ in terms of the gamma function for any $k>0$. But following the definition $M_X(t)=\int_{-\infty}^\infty e^{tx}f(x)dx$ is very complicated here, since I have to first compute the probability distribution function, and then integrate a very complicated function. I don't think it's the way to find $E[X^k]$. What would be the way to do it?

The probability density function is $-\frac{d}{dt} P(X>t)$. Hence 
$$
E[X^k] =  - \int_{0}^\infty t^k\, d(P(X>t)) =  \int_{0}^\infty P(X>t)\,d(t^k)
=k \int_{0}^\infty t^{k-1} e^{-ct^\beta}\,dt
$$
Change the variable to $s=ct^\beta$, which makes  $ds=c\beta t^{\beta-1}\,dt$:
$$
E[X^k] = \frac{k}{c\beta}   \int_{0}^\infty t^{k-\beta} e^{-s}\,ds =
\frac{k}{c^{k/\beta} \beta}   \int_{0}^\infty s^{k/\beta-1} e^{-s}\,ds = \frac{k}{c^{k/\beta} \beta} \Gamma(k/\beta)$$
