Raw Moment of Weibull Distribution

Theorem
Let $X$ be a continuous random variable with the Weibull distribution with $\alpha, \beta \in \R_{> 0}$.

Let $n$ be a strictly positive integer.

Then the $n$th raw moment $\expect {X^n}$ of $X$ is given by:


 * $\ds \expect {X^n} = \beta^n \map \Gamma {1 + \frac n \alpha}$

where $\Gamma$ is the Gamma function.

Proof
From the definition of the Weibull distribution, $X$ has probability density function:


 * $ \map {f_X} x = \alpha \beta^{-\alpha} x^{\alpha - 1} e^{-\paren {\frac x \beta}^\alpha}$

Where $\Img X = \R_{\ge 0}$.

From the definition of the expected value of a continuous random variable:
 * $\ds \expect {X^n} = \int_0^\infty x^n \map {f_X} x \rd x$

Therefore:

Let:
 * $u = \paren {\dfrac x \beta}^\alpha$

Then by Chain Rule for Derivatives, we have:
 * $\rd u = \alpha \paren {\dfrac x \beta}^{\alpha - 1} \dfrac 1 \beta \rd x$

And also:
 * $\paren {\beta u^{\frac 1 \alpha} }^n = x^n$

We can see that as $x \to 0$, $u \to 0$ and as $x \to \infty$, $u \to \infty$

Plugging these results back into our integral above, we have: