Let $X$ be $N(0,1)$. Show that $E\{X^{2n+1}\}=0$ (Easy - calculate it directly using the definition of expectation, and you're taking the integral of an odd function over a symmetric interval, so =0), and $E\{X^{2n}\}=\frac{(2n)!}{2^{n}n!}=(2n-1)(2n-3)\cdots 3\cdot 1$ (what I'm having trouble with).
I'm really struggling to understand how to do this one - Please help.
Since $X\sim \mathcal{N}(0,1)$, it's pdf is $f(x)=\frac{1}{\sqrt{2\pi}} \exp\left(-\frac{1}{2}x^2\right)$.
Now, it follows that
$$E[X^{2n}] = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} x^{2n}\exp\left(-\frac{1}{2}x^2\right)\,dx = \frac{2}{\sqrt{2\pi}} \int_0^{\infty} x^{2n}\exp\left(-\frac{1}{2}x^2\right)\,dx $$
Since the integrand is an even function. Now make the substitution $t=\frac{1}{2}x^2 \implies \,dt = x\,dx$
Hence, we now see that
$$\begin{aligned} E[X^{2n}] &= \frac{2}{\sqrt{2\pi }} \int_0^{\infty} (2t)^{n-1/2} \exp (-t)\,dt \\ &= \frac{2^{n+1/2}}{\sqrt{2\pi}}\int_0^{\infty}\exp(-t) t^{n-1/2}\,dt \\ &= \frac{2^n}{\sqrt{\pi}}\Gamma\left(n+\frac{1}{2}\right)\end{aligned}$$
Now note that
$$\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n-1)(2n-3)\cdots(3)(1)}{2^n}\Gamma\left(\frac{1}{2}\right) = \frac{(2n-1)(2n-3)\cdots 3\cdot 1}{2^n}\sqrt{\pi}$$
And thus
$$\begin{aligned} E[X^{2n}] &= (2n-1)(2n-3)\cdots 3\cdot 1\\ &=\frac{(2n)(2n-1)(2n-2)(2n-3)\cdots (3)(2)(1)}{(2n)(2n-2)\cdots(4)(2)}\\ &= \frac{(2n)!}{2^n n!}\end{aligned}$$