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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.

saikat
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    You can just use binomial expansion and integrate term by term. –  Apr 13 '15 at 06:19
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    The change of variable $t=x^p$ yields directly the (quite complete) Beta function $$\tfrac1p\cdot\mathrm{Beta}\left(n+1,\tfrac1p\right).$$ – Did Apr 13 '15 at 06:30

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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:

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

Jack D'Aurizio
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  • thanks for that sweet answer. I just wanted to know, when you make the assumption, $p>0$ and $n \in N $\ {0}, in which step do make those assumptions? – saikat Apr 13 '15 at 14:29
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    @saikat: $n\in\mathbb{N}\setminus{0}$ is needed to write the RHS of $(2)$ as such, while $p>0$ grants that the integral in the middle term of $(1)$ is converging. – Jack D'Aurizio Apr 13 '15 at 17:30
  • As a continuation, if we have the integral limits changed to [0,$\infty$], i.e., $\int_{0}^{\infty}(1-x^p)^ndx$, then how does the definite integral change? – saikat Apr 16 '15 at 18:55
  • @saikat: the integral is diverging if the integration range is extended to $(0,+\infty)$. – Jack D'Aurizio Apr 16 '15 at 18:56
  • Does that mean that it would go to $\infty$ for any value of $n$ and $p$? – saikat Apr 16 '15 at 19:07
  • @saikat: assuming $p>0$ and $n\in\mathbb{N}$, yes. – Jack D'Aurizio Apr 16 '15 at 19:10
  • got it. Just one last question. How do you derive: $\frac{1}{p}\int_{0}^{1}z^{\frac{1}{p}-1}{(1-z)}^ndz = \frac{1}{p}\frac{\Gamma(\frac{1}{p})\Gamma(n+1)}{\Gamma(n+1+\frac{1}{p})}$ ? – saikat Apr 16 '15 at 19:17
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    @saikat: http://en.wikipedia.org/wiki/Beta_function – Jack D'Aurizio Apr 16 '15 at 19:19