If $X$~WEI$(\theta,\beta)$, derive $E(X^k)$ assuming $k\gt-\beta$.
Note that $X$~WEI$(\theta,\beta)=\frac{\beta}{\theta^{\beta}}x^{\beta -1}e^{-({x}/{\theta})^{\beta}}$
I am having a very difficult time attempting this problem. I start with:
$$E(X^k)=\int_0^{\infty}x^k \frac{\beta}{\theta^{\beta}}x^{\beta -1}e^{-({x}/{\theta})^{\beta}}dx$$
$$=\frac{\beta}{\theta^{\beta}}\int_0^{\infty}x^{\beta+k-1}e^{-({x}/{\theta})^{\beta}}dx$$
It was hinted to me to attempt substitution, so I let $({x}/{\theta})^{\beta}=t$
Based on a similar problem I did, I have the suspicion that I need to get it in the form of:
$$\int_0^{\infty}t^{k-1}e^{-t}dt$$ which is the gamma function.
With the substitution $({x}/{\theta})^{\beta}=t$ we quickly get the $e^{-t}$ portion. I'm having trouble finding the steps to make the connection between the rest of the problem and $t^{k-1}$. Or, is this not the correct approach? Any help would be greatly appreciated!
You can calculate $\int_0^{\infty}x^k \frac{\beta}{\theta^{\beta}}x^{\beta -1}e^{-({x}/{\theta})^{\beta}}dx$ using integration by parts $k$ times, if you do it you finally get $\int_{0}^{\infty} \frac{\beta}{\theta^{\beta}}x^{\beta -1}e^{-({x}/{\theta})^{\beta}} dx$ and you know that it's equals $1$.