Statistical Integral in Financial Mathematics

Question

I need to show that $$\frac{1}{\sqrt {2\pi}} \int_{-\infty}^{+\infty} x^{2n} e^{\frac{-x^2}{2}}dx = (2n-1)!!$$ Integration by parts seems to be the best apporach but I cannot seem to figure my way through it.

Answer 1

$$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x^{2n}e^{-\frac{x^2}{2}}dx=\sqrt{\frac{2}{\pi}}\int_0^\infty x^{2n}e^{-\frac{x^2}{2}}dx$$ Let $u=\frac{x^2}{2}$, $dx=\frac{du}{\sqrt{2u}}$ $$I=\frac{2^n}{\sqrt{\pi}}\int_0^\infty u^{n-\frac{1}{2}}e^{-x}dx=\frac{2^n}{\sqrt{\pi}}\Gamma(n+\frac{1}{2})$$ Using $\Gamma(\frac{1}{2})=\sqrt{\pi}$ and $\Gamma(n+1)=n\Gamma(n)$ it is easy to show that $$\Gamma(n+\frac{1}{2})=\frac{(2n-1)!!\sqrt{\pi}}{2^n}$$ So $$I=\frac{2^n}{\sqrt{\pi}}\Gamma(n+\frac{1}{2})=\frac{2^n}{\sqrt{\pi}}\frac{(2n-1)!!\sqrt{\pi}}{2^n}=(2n-1)!!$$

We use the fact that $$\Gamma(n+1)=n!=\int_0^\infty x^ne^{-x}dx$$ and $$\Gamma(\frac{1}{2})=\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$$

Answer 2

Normally I'd prove this by differentiation, but Zachary already mentioned that, so here's an alternative. Call this integral $I_n$ so $$I_{n+1}-(2n+1)I_n=\frac{1}{\sqrt{2\pi}}\int_{\Bbb R}\bigg(x^{2n+2}\exp-\frac{x^2}{2}-(2n+1)x^{2n}\exp-\frac{x^2}{2}\bigg)dx\\=\frac{1}{\sqrt{2\pi}}\bigg[-x^{2n+1}\exp-\frac{x^2}{2}\bigg]_{-\infty}^\infty=0.$$