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

 A: \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.)
A: 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}
