If $X$~$\text{Norm}(0,\sigma^2)$ find $\Bbb E[X^{2n}]$, where $n$ is a natural number.
I can work the integral up to a point: $$\int x^{2n} \frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-x^2}{2\sigma^2}}dx$$ Change of variable: $v = \frac x\sigma$ $$\to \int (\sigma v)^{2n} \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dv$$ $$=\sigma^{2n}\int v^{2n} \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dv$$ Here's where I get lost: I think the $v^{2n} $needs to be gotten rid of. How would I do it?
Integration by parts here will probably work, but is long. A quicker solution is to note the the moment generating function is $$ M(t) = e^{\frac{1}{2} \sigma^2 t^2} = \sum_{k=0}^{\infty} \frac{1}{k!} \frac{\sigma^{2k} }{2^k} t^{2k} = \sum_{k=0}^{\infty} \frac{1}{(2k)!} \frac{\sigma^{2k}(2k)! }{2^k(k!)} t^{2k} $$ Hence $$ E[X^{2k}] = \frac{\sigma^{2k}(2k)! }{2^k(k!)} $$
It suffices to consider $X\sim N(0,1)$ since if $W\sim N(0,\sigma ^2)$, it follows that $W=\sigma Z$ where $Z\sim N(0,1)$ and $EW^k=\sigma^k Z^k$. To this effect we must compute $$ EX^{2n}=\int_{-\infty}^\infty\frac{x^{2n}}{\sqrt{2\pi}}e^{-x^2/2}\,dx.\tag{0} $$ Define $$ F(t)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-tx^2/2}\,dx\quad(t>0) \tag{1} $$ and we compute $F^{(n)}(1)$ in two different ways. First by differentiation under the integral sign $$ F^{(n)}(t)=\frac{(-1)^n}{2^n}\int_{-\infty}^\infty\frac{x^{2n}}{\sqrt{2\pi}}e^{-tx^2/2}\,dx\implies F^{(n)}(1)=\frac{(-1)^n}{2^n}EX^{2n}.\tag{2} $$ Second make the substitution $u=x(t/2)^{1/2}$ in (1) and observe that $$ F(t)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{{2}}{t}}\underbrace{\int_{-\infty}^\infty e^{-u^2}\, du}_\text{$=\sqrt{\pi}$}=t^{-1/2} \implies F^{(n)}(1)=\frac{(-1)^n}{2^n}(1\cdot3\cdot5\cdots2n-1) \tag{3} $$ Comparing (2) and (3) it follows at once that $$ EX^{2n}=1\cdot3\cdot5\cdots2n-1=\frac{(2n)!}{2^nn!}. $$
