Expectation of normal distributions

Question

Let $X$ have a normal distribution with mean $µ$ and variance $σ^2$.
Find $E[X^3]$ (in terms of $µ$ and $σ^2$).

the pdf of this function is $$\frac{1}{σ\sqrt{2\pi}} e^{\frac{-(x-µ)^2}{2σ^2}}$$

so $$E[X^3] = \int_{-\infty}^\infty x^3\frac{1}{σ\sqrt{2\pi}} e^{\frac{-(x-µ)^2}{2σ^2}} dx$$
How do I compute this integral?

Answer 1

Using the symmetry of $\frac{X-\mu}{\sigma}$, one can conclude that $$ E\left(\frac{X-\mu}{\sigma}\right)^{2n+1}=0, \text{for}\quad n=0,1,2,... $$ Hence, using $Var(X)=\sigma^2$ and $EX=\mu$, you can show that $E\left(\frac{X-\mu}{\sigma}\right)^{3}=\frac{EX^3-3\mu\sigma^2-\mu^3}{\sigma^3}=0$, thus $$ EX^3 = 3\mu\sigma^2 + \mu^3. $$