Integral hangs G2

[Albert Chan]

(03-29-2020 09:56 PM)Albert Chan Wrote:  Googled "wallis product", I found this: Mathematical Analysis and the Mathematics of Computation, p.404, eqn. 7.87

\(\Large {1 \over \int_0^1 (1-x^{1/p})^q\;dx} = \binom{p+q}{p} = \binom{p+q}{q} = {1 \over \int_0^1 (1-x^{1/q})^p\;dx} \)

Just realized the integral is simply the beta function (see equation 18)

Let \((1-u)^p = x\quad →\quad -p(1-u)^{p-1}\;du=dx\)

\(\large \int_0^1 (1-x^{1/p})^q\;dx
= p \int_0^1 u^q (1-u)^{p-1}\;du
= p\;B(q+1,p)
= {\Gamma(p+1)\Gamma(q+1) \over \Gamma(p+q+1)}
= 1 / \binom{p+q}{p} \)

Update:

We can prove u integral is beta function by induction (via integration by parts)
Foundations of Combinatorics with Applications:       Appendix A, Example A.4