I have to solve the integral $\int_{-\infty}^{\infty} N(x; \mu, \sigma)^p\,dx$. I remember I had to do this before and came up with an easy and elegant solution, but just can't remember how I got there and it's been a while. Could someone help me out?
Thanks a lot in advance!
I assume that $N(x;\mu,\sigma)$ is the density of the normal distribution. I also assume that $p$ is positive. In that case, $$N(x;\mu,\sigma)=\frac{1}{\sigma\sqrt{2\pi}}\mathrm{e}^{-\frac{(x-\mu)2}{\sigma^2}},$$ so that $$N(x;\mu,\sigma)^p=\frac{1}{\sigma^p(2\pi)^{p/2}}\mathrm{e}^{-p\frac{(x-\mu)2}{\sigma^2}}=\frac{\frac{\sigma}{\sqrt{p}}\sqrt{2\pi}}{\sigma^p(2\pi)^{p/2}}N(x;\mu,\frac{\sigma}{\sqrt{p}}).$$ Integrating this expression, you get $$\int_{\mathbb{R}}N(x;\mu,\sigma)^pdx=\frac{\frac{\sigma}{\sqrt{p}}\sqrt{2\pi}}{\sigma^p(2\pi)^{p/2}}=\frac{(\sigma \sqrt{2\pi})^{1-p}}{\sqrt{p}}.$$
