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**approx** I've got this problem:

f(y) = $\displaystyle (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}$, y>0

0, elsewhere

Now I'm supposed to find $\displaystyle E(Y^k)$ for any positive integer k.

I start out like this:

$\displaystyle E(Y^k)$ = $\displaystyle \int_0^\infty y^{k}f(y) dy$ = $\displaystyle \int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy$

I then move out $\displaystyle (1/\alpha)$ and m outside the integral =

$\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy$ = $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$

and from here I don't know how to go on. The correct answer is $\displaystyle \Gamma (k/m+1)\alpha ^{k/m}$

and I know that $\displaystyle \Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy$

but I don't see how $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$ can be rewritten to the correct answer. Thanks in advance for your help.