How to find the third moment $E[X^3]$ of Gamma Distribution?

Question

I'm having trouble with the following question. So far I've tried to integrate $$\int_{0}^{\infty} x^3f(x) \, dx$$ but I end up with a horrible looking integral (I assume integration by parts won't fix this...)

Answer 1

Sketch:

\begin{align} E[X^3] &= \int_0^\infty x^3 f_X(x) dx \\ &=\int_0^\infty \frac{\lambda^\alpha x^{3+\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)} dx \\ &= \frac{1}{\lambda^3}\int_0^\infty \frac{\lambda^{\alpha + 3} x^{3+\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)} dx \\ &= \frac{\Gamma(\alpha+3)}{\lambda^3 \Gamma(\alpha)}\int_0^\infty \frac{\lambda^{\alpha + 3} x^{3+\alpha-1}e^{-\lambda x}}{\Gamma(\alpha+3)} dx \\ \end{align}

Answer 2

\begin{align} & \int_0^\infty x^{\alpha - 1} e^{-\lambda x} \, dx = \frac {\Gamma(\alpha)} {\lambda^\alpha}. \tag 1 \\[15pt] \text{Therefore } & \int_0^\infty x^3 \cdot x^{\alpha-1} e^{-\lambda x} \\[10pt] = {} & \int_0^\infty x^{(\alpha+ 3)-1} e^{-\lambda x} \, dx \\[10pt] = {} & \text{the same thing as (1) except that } \alpha-3 \\ & \qquad \text{appears where } \alpha \text{ had been, and therefore} \\[12pt] = {} & \frac {\Gamma(\alpha+3)} {\lambda^{\alpha+3}}. \end{align}

(Explicit point about logic: The equality $(1)$ is true if $\alpha$ is any positive number at all. Since $\alpha+3$ is a positive number, it is true if $\alpha+3$ appears where $\alpha$ had been.)

So first use that in evaluating the relevant integral.

Then use this: $\Gamma(\alpha+3) = (\alpha+2)(\alpha+1)\alpha\Gamma(\alpha).$

(This same sort of thing applies to finding moments of the Beta distribution.)