I need to find the value of the constant $a$ such that $af^n$ is a pdf given that $f$ is a normal pdf with parameters $\mu$ and $\sigma^2$.
My attempt: I tried to look at the integral of the normal pdf over the real line, multiply it by $a$, and set it equal to $1$. But I didn't know what to do with the $n$ in the exponent.
The pdf of the variable $X\sim\mathcal{N}(\mu,\sigma^2)$ can be expressed in terms of the standard normal pdf $\phi(x)$ as follows:
$$f_X(x)=\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)$$ So we have
$$af_X(x)^n=\frac{1}{\sigma^n}\phi\left(\frac{x-\mu}{\sigma}\right)^n=\frac{1}{\sigma^n}\phi\left(\frac{1}{\sigma}x-\frac{\mu}{\sigma}\right)^n$$ in order to be pdf, you should have $$\int_{-\infty}^{+\infty}af_X(x)^ndx=1$$ You can use this to find the integral and solve the above equation for $a$.