1. ## Wallis' integral

How would you show that $\displaystyle{ \int_0^1 (x-x^2)^n dx = \frac{(n!)^2}{(2n+1)!} }$ ?

2. Might not be much help but,

$\displaystyle \int_0^1 (x-x^2)^n ~dx = \int_0^1 (x(1-x))^n ~dx= \int_0^1 x^n(1-x)^n ~dx = \frac{\Gamma (n)\Gamma (n)}{\Gamma (2n)}= \frac{(n-1)!(n-1)!}{(2n-1)!}$

Now have a play around with those factorials.

3. Originally Posted by pickslides
Might not be much help but,

$\displaystyle \int_0^1 (x-x^2)^n ~dx = \int_0^1 (x(1-x))^n ~dx= \int_0^1 x^n(1-x)^n ~dx = \frac{\Gamma (n)\Gamma (n)}{\Gamma (2n)}= \frac{(n-1)!(n-1)!}{(2n-1)!}$

Now have a play around with those factorials.
The 'Beta function' is...

$\displaystyle B(h,k)= \int_{0}^{1} x^{h-1}\ (1-x)^{k-1}\ dx$ , $h>0, k>0$ (1)

... and its 'connection' with the 'Gamma function' is...

$\displaystyle B(h,k)= \frac{\Gamma(h)\ \Gamma(k)}{\Gamma(h+k)}$ (2)

Setting n=h+1=k+1 in (1) and (2) we obtain...

$\displaystyle \int_{0}^{1} x^{n}\ (1-x)^{n}\ dx = \frac{\Gamma(n+1)\ \Gamma(n+1)}{\Gamma(2n+2)} = \frac{(n!)^{2}}{(2n+1)!}$ (3)

Kind regards

$\chi$ $\sigma$