What is $\int\limits_{0}^{1} \left[x(1-x)\right]^m \, dx$ ($m$ positive integer)?

Question

I came across the following integral in my research:

$$ \int\limits_{0}^{1} \left[x(1-x)\right]^m \, dx \qquad m\in\mathbb{N}^+ $$

According to my CAS (I use Matlab's Symbolic Toolbox), this evaluates to:

$$ \frac{\left(m!\right)^2}{\left(2m + 1\right)!} $$

Checking a few low values of $m$ suggests that this result is correct, but I cannot understand how it was derived.

I have tried expanding $x(1-x) = x - x^2$ and using the Binomial theorem. This seems to be a step in the right direction (at least it introduces factorials), but I am not managing to grind the algebra to get that result.

Answer 1

As Cameron Williams commented, this is related to the beta function.

Making it more general $$\int x^n (1-x)^m\,dx=B_x(n+1,m+1)$$ and $$\int_0^1 x^n (1-x)^m\,dx=\frac{\Gamma (m+1)\, \Gamma (n+1)}{\Gamma (m+n+2)}=\frac{m! \, n! }{(m+n+1)!}$$

Answer 2

The binomial theorem works. Write $$ \begin{align} \int_0^1(x-x^2)^m\,dx&=\int_0^1\sum_{k=0}^m{m\choose k}x^{m-k}(-x^2)^k\,dx\\ &=\sum_{k=0}^m{m\choose k}(-1)^k\int_0^1x^{k+m}\,dx\\ &=\sum_{k=0}^m{m\choose k}(-1)^k\frac1{k+m+1}\,dx\\ \end{align} $$ Now the tricky part: apply the identity $$\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{k+x} = \frac{1}{x\binom{m+x}{m}} $$ with $x=m+1$ to see this is equal to $$\frac1{(m+1){2m+1\choose m}}={(m!)^2\over(2m+1)!}.$$

Answer 3

Hint:

Let $I_{m,m} = \int_0^1 \! x^m(1-x)^m \,dx$ , for $m\in\mathbb{N}^+$

$I_{m,m} = \int_0^1 \! x^m(1-x)^m \,dx = \int_0^1 x^m \,\frac{d(1-x)^{m+1}}{-(m+1)}$

Using Integration by Parts,

$= \frac{-1}{m+1}[x^m(1-x)^{m+1}]_0^1 + \frac{1}{m+1}\int_0^1 \! (mx^{m-1})(1-x)^{m+1} \,dx $

$= \frac{m}{m+1} I_{m-1,m+1}$

Hence, you could observe the recurrence form