Curious Integrals

This problem can be tricky. The most direct way is to start by making the substitution $u = 1 - {x}^{1/p}$ and $v = 1 - {x}^{1/q}$ . Then evaluate the two integrals using the Beta function. Since the Beta function is commutative, $A=B;$ therefore, $A - B = 0.$

The answer is actually not independent of p and q. It can be shown that A=B for all positive values of p and q, and that both integrals equal: (p!)(q!)/(p+q)!

For A: Let $u=1-x^{1/p}$ $\Rightarrow$ $du=-\frac{1}{p}\frac{x^{\frac{1}{P}}}{x}dx$

$\Rightarrow$ $dx=-p(1-u)^{p-1}du$

$\Rightarrow$ $A=\int^0_1 (-pu^q(1-u)^{p-1})\,du$

Integrating by parts yields: $A=p(\frac{u^q(1-u)^p}{p}+\frac{qu^{q-1}(1-u)^{p+1}}{p(p+1)}+...+\frac{q!(p-1)!(1-u)^{p+q-1}}{(p+q)!})$ Which we evaluate from 1 to 0. When $u=0$ , it is clear that all terms except the last drop out. When $u=1$ all terms will drop out, hence we have:

$A=\frac{q!p!}{(p+q)!}$

B is evaluated in the exact same way, with v substituted for u.

Also the answer is independent of p and q ... !! a troll problem anyways :P