$E[X]$ for $X \sim$ Weibull $(\alpha)$

Question

I am trying to evaluate the expectation of a Weibull distribution whose form is

$$f_X(x;\alpha)=\alpha x^{\alpha-1}e^{-x^\alpha} $$

So, I am trying to evaluate the intergral

$$E[X] = \int_0^{\infty} \alpha x^{\alpha} e^{-x^{\alpha}}dx$$

but I am having trouble.

u-substitution causes $x^{\alpha-1}$ to be in the way, and integration by parts makes $e^{-x^\alpha}dx$ a tough expression to deal with.

I have also considered expressing the expectation as

$$\frac{1}{\alpha +1}\int_0^{\infty} (\alpha +1)x^\alpha e^{-x^{\alpha +1}}*e^{-x^{-1}}dx $$

to utilize

$$\int_0^{\infty} (\alpha +1)x^\alpha e^{-x^{\alpha +1}}dx =1 $$

but am not able to manage things well with the extra $e^{-x^{-1}}$.

I do not have the strongest real analysis skill since I have never reached that level, but if this takes some knowledge above highschool level calculus, I would at least like to know.

Thank you.

Answer 1

The answer is well-known; see https://en.wikipedia.org/wiki/Weibull_distribution

If you just need the answer for an applied problem that will do it. I have not been able to find a derivation, however. The section on deriving the MGF and moments in that reference may be of some help.