how to compute the integral $\int_0^1 (1-x^p)^n dx$?

Question

For constants $n$ and $p$, how to compute the integral $\int_0^1 (1-x^p)^n dx$ ?

I saw a solution using hypergeometric function and another using incomplete beta function here: http://www.wolframalpha.com/input/?i=integrate+%281-x^p%29^n+dx

However, I can't find the steps to approach this problem and also unable to compute the definite integrals from 0 to 1.

Answer 1

Just expanding Did's comment, $$I=\int_{0}^{1}(1-x^p)^n\,dx = \frac{1}{p}\int_{0}^{1}z^{\frac{1}{p}-1}(1-z)^n\,dz = \frac{1}{p}\cdot\frac{\Gamma\left(\frac{1}{p}\right)\Gamma(n+1)}{\Gamma\left(n+1+\frac{1}{p}\right)}\tag{1}$$ and by using the identity $\Gamma(z+1)=z\,\Gamma(z)$ multiple times we get:

$$ I = \frac{n!}{\prod_{k=1}^{n}\left(k+\frac{1}{p}\right)}=\prod_{k=1}^{n}\left(1+\frac{1}{pk}\right)^{-1}.\tag{2}$$

We may also achieve the same result by substituting $z=x^p$ then applying integration by parts multiple times. Just to be clear, I am assuming $p>0$ and $n\in\mathbb{N}\setminus\{0\}$. By the symmetry of the RHS of $(1)$ we also have:

$$ \int_{0}^{1}(1-x^p)^n\,dx = \int_{0}^{1}(1-x^{1/n})^{1/p}\,dx. \tag{3}$$