The question is:
Show that if $X ∼ N(0, 1)$ has the standard normal distribution then
$E[X^{2n}] = \frac{2n!}{2^{n}n!}$
Hint: compute the integral $\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-sx^{2}} dx$ and take the derivative of both sides with respect to $s$.
To say I'm stuck would be an understatement... Thanks for any help
Let $f(s)=\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-sx^{2}} dx$. The substitution $u=\sqrt{s}x$ together with the standard Gaussian integral gives that $f(s)=\sqrt{\frac{1}{2s}}$. Now, we take the expression:
$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-sx^{2}} dx=\sqrt{\frac{1}{2s}}$$
and differentiate both sides $n$ times with respect to $s$:
$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-sx^{2}}(-x^2)^n dx=\frac{1}{\sqrt{2}{s^{n+1/2}}}\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right) \cdots \left(\frac{-(2n-1)}{2}\right)$$
Setting $s=\frac{1}{2}$, we see that
$$(-1)^nE(X^{2n})=\frac{1}{\sqrt{2}\left(\frac{1}{2}\right)^{n+1/2}}\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right) \cdots \left(\frac{-(2n-1)}{2}\right)=(-1)^n{1\cdot3\cdots(2n-1)}$$
Rewriting $1\cdot3\cdots(2n-1)=\frac{1\cdot2\cdots(2n)}{2\cdot4\cdots(2n)}=\frac{(2n)!}{2^nn!}$, we arrive at the desired result.