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 $ds=c\beta t^{\beta-1}\,dt$:
$$
E[X^k] = \frac{k}{c\beta} \int_{0}^\infty t^{k-\beta} e^{-s}\,ds =
\frac{k}{c^{k/\beta} \beta} \int_{0}^\infty s^{k/\beta-1} e^{-s}\,ds = \frac{k}{c^{k/\beta} \beta} \Gamma(k/\beta)$$
I found explicit formulae using the generalized incomplete Beta function defined as follows (this is a standard special function available in numerical packages):
$$\mathrm{B}\left(z_{1},z_{2};\alpha,\beta\right)=\int_{z_{1}}^{z_{2}}x^{\alpha-1}\left(1-x\right)^{\beta-1}\mathrm{d}x$$
It follows that
$$g(x) = \frac{1}{Z}\left(x-A\right)^{\alpha-1}\left(B-x\right)^{\beta-1}$$
with the normalization constant
$$Z=\int_{a}^{b}\left(x-A\right)^{\alpha-1}\left(B-x\right)^{\beta-1}\mathrm{d}x
=\left(B-A\right)^{\alpha+\beta-1}\mathrm{B}\left(a,b;\alpha,\beta\right)$$
Instead of computing $\left\langle x\right\rangle$ and $\langle x^2\rangle$ directly, we compute $\langle x-A\rangle$ and $\langle (x-A)^{2}\rangle$. We have:
$$\left\langle x-A\right\rangle =\frac{1}{Z}\int_{a}^{b}\left(x-A\right)^{\alpha}\left(B-x\right)^{\beta-1}\mathrm{d}x
=\frac{1}{Z}\left(B-A\right)^{\alpha+\beta}\mathrm{B}\left(a,b;\alpha+1,\beta\right)
=(B-A)\frac{\mathrm{B}\left(a,b;\alpha+1,\beta\right)}{\mathrm{B}\left(a,b;\alpha,\beta\right)}$$
and, similarly:
$$\langle \left(x-A\right)^{2}\rangle =\left(B-A\right)^{2}\frac{\mathrm{B}\left(a,b;\alpha+2,\beta\right)}{\mathrm{B}\left(a,b;\alpha,\beta\right)}$$
From these it is easy to get $\langle x \rangle$ and $\langle x^2 \rangle - \langle x \rangle^2$:
$$\langle x \rangle = \langle x - A \rangle + A$$
$$\langle x^2 \rangle - \langle x \rangle^2 = \langle (x-A)^2 \rangle - \langle x-A \rangle^2$$
Perhaps some simplifications are possible, but I do not know them. If someone can help on that it would be awesome.
Best Answer
For constant $k$, we have the following $$E(X^k)=\int\alpha x^{\alpha+k-1}e^{-x^\alpha}dx$$ Using substitution $u=x^\alpha\Leftrightarrow x=u^{1/\alpha}$ results in $$du=\alpha x^{\alpha-1}dx \Rightarrow dx=\frac{du}{\alpha x^{\alpha-1}}=\frac{du}{\alpha u^{\frac{\alpha-1}{\alpha}}}=\left(u^{\frac{1}{\alpha}-1}\right)\frac{du}{\alpha}$$ leading to $$\begin{align}E(X^k)&=\int\alpha \color{blue}{x^{\alpha+k-1}}\color{red}{e^{-x^\alpha}}\color{green}{dx}\\&=\int\alpha \color{blue}{u^{(\frac{k}{\alpha}+1)-\frac{1}{\alpha}}}\color{red}{e^{-u}}\color{green}{\left(u^{\frac{1}{\alpha}-1}\right)\frac{du}{\alpha}}\\&=\int u^{\left(\frac{k}{\alpha}+1\right)-1}e^{-u}\ du\\&=\Gamma\left(\frac{k}{\alpha}+1\right)\end{align}$$ Thus the mean is $$E(X)=\Gamma\left(\frac{1}{\alpha}+1\right)$$ and the variance is $$Var(X)=E(X^2)-(E(X))^2=\Gamma\left(\frac{2}{\alpha}+1\right)-\Gamma^2\left(\frac{1}{\alpha}+1\right)$$