<s>[INST] Expectation of a random variable Suppose $Y$ is normal random variable. 
Show that for any integer $n \ge 1$,  
$$
E[(Y- \mu)^{2n}] = \sigma^{2n} \frac{2n!}{2^{n}n!}.
$$  
I've mangaged to show that that $f(\mu+y)$ = $f(\mu-y)$ and that therefore $(Y-\mu)$ and $(\mu - Y)$ have the same distribution. I think I have to somehow use this information to help compute the expectation, but I still do not know how to deal with the integral.
Can someone offer some help?
 [/INST] Assume $Y\sim N(\mu,\sigma)$ is a random variable with a normal distribution. Then for any $n\geq 1$, 
$$
E[(Y-\mu)^{2n}] = E\left[\left(\sigma\cdot \dfrac{Y-\mu}{\sigma}\right)^{2n}\right] =\sigma^{2n}E\left[\left( \dfrac{Y-\mu}{\sigma}\right)^{2n}\right]. 
$$
Let $X=\dfrac{Y-\mu}{\sigma}$. Then $X\sim N(0,1)$ has the standard normal distribution.  So we need to calculate 
$$
E[X^{2n}]=E\left[\left( \dfrac{Y-\mu}{\sigma}\right)^{2n}\right].
$$  
Since 
$$
\begin{aligned}
\left( \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-cx^2} dx\right)^2 &= 
 \int_{-\infty}^{\infty}  \int_{-\infty}^{\infty} \dfrac{1}{2\pi} e^{-c(x^2+y^2)}dx\, dy \\
&= \int_{0}^{2\pi}  \int_{0}^{\infty}\dfrac{1}{2\pi} e^{-c r^2}r\, dr\, d\theta \\
&= \frac{1}{2} \int_{0}^{2\pi}  \int_{0}^{\infty}\dfrac{1}{2\pi} e^{-c u}du \, d\theta \hspace{4mm}\mbox{ where }u=r^2\\
&= \dfrac{1}{2}\dfrac{2\pi}{2\pi}\left( -\frac{1}{c}\right) e^{-cu}\Big|_{u=0}^{\infty} \\ 
&= \left( -\dfrac{1}{2c} \right) \left( 0-1\right) =\dfrac{1}{2c}, 
\end{aligned}
$$
we conclude
\begin{equation}
 \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-cx^2} dx = \sqrt{\dfrac{1}{2c}}.  \hspace{16mm} (\dagger) 
\end{equation} 
This is called the $\textbf{Gaussian integral}$, whose technique is often used in mathematical physics and complex analysis. 
Now, take the derivative of both sides of $(\dagger)$ with respect to $c$ $n$-times to obtain: 
$$ 
\int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}}\left( -x^2 \right)^n e^{-cx^2}dx 
= \dfrac{1}{\sqrt{2}}\left(-\dfrac{1}{2} \right) \left(-\dfrac{3}{2} \right) \cdots \left(-\left( \dfrac{2n-1}{2}\right) \right) c^{-\frac{2n+1}{2}}.
$$
Now let $c=\frac{1}{2}$. Then 
$$
\begin{aligned}
(-1)^n E[X^{2n}] &= 2^{\frac{2n+1}{2}-\frac{1}{2}}(-1)^n \dfrac{1}{2} \dfrac{3}{2} \cdots \dfrac{2n-1}{2} \\ 
&= 2^n (-1)^n \dfrac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^n} \\ 
&=  (-1)^n 1\cdot 3 \cdot 5 \cdots (2n-1)  \\ 
&= (-1)^n \dfrac{1\cdot 2\cdot 3 \cdot 4\cdots 2n}{2\cdot 4\cdot 6 \cdot 8\cdots 2n} \\ 
&= (-1)^n \dfrac{(2n)!}{2^n(1\cdot 2 \cdot 3\cdot 4\cdots n)} \\ 
&= (-1)^n \dfrac{(2n)!}{2^n \cdot n!}. 
\end{aligned}
$$
Since 
$$
E[X^{2n}] = \dfrac{(2n)!}{2^n \cdot n!}, 
$$
we conclude that 
$$
E[(Y-\mu)^{2n}] = \sigma^{2n} \dfrac{(2n)!}{2^n \cdot n!}. 
$$
</s>