Wikipedia:Reference desk/Archives/Mathematics/2014 June 16

= June 16 =

Evaluating an Integral
How would one prove that $$\int_0^\infty\bigg[\Big(1+\sqrt[a]x\Big)^a-x\bigg]dx~=~\frac12\cdot{-a\choose a}^{-1}$$ ? — GX, May 1971 (talk) 22:00, 16 June 2014 (UTC)
 * I'm not sure about the notation on the right side of your equation (do you mean 1/(-a C a)?), but my first thought is the binomial theorem followed by term-by-term integration. I have this hunch that it's related to the gamma function or beta function.--Jasper Deng (talk) 00:56, 17 June 2014 (UTC)
 * By doing that, I was able to re-write the definite integral as a convergent infinite series, but how this will help in solving the actual problem is beyond me.
 * $$\frac12+a\sum_{n=1}^\infty{a\choose n}\Bigg(\frac1{n+a}+\frac1{n-2a}\Bigg)=\frac12\cdot{-a\choose a}^{-1}~-$$ GX, May 1971 (talk) 01:43, 17 June 2014 (UTC)
 * Hhm... after finding that this would fall flat on its face at first, I then tried the substitution $$u^a=x$$ which converts the integral to $$\int_0^\infty ((1+u)^a-u^\frac{1}{a})au^{a-1} du$$. This kinda resembles the beta function. I'll investigate further later on this evening.--Jasper Deng (talk) 02:18, 17 June 2014 (UTC)
 * It should be $$u^a$$ there, not $$u^\frac1a\quad$$ — GX, May 1971 (talk) 02:38, 17 June 2014 (UTC)
 * After further investigation (thanks for that correction, by the way), I use the substitution $$v=u+1$$ to obtain $$\int_{-1}^\infty (v^a-(v-1)^a)a(v-1)^{a-1} dv$$. Since $$a$$ and $$(-1)^a$$ are constant, I finally get an expression in terms of the incomplete beta function, $$a(-1)^{(a-1)}(\Beta(-1; a+1,a)+\Beta(\infty; a+1, a)) - a(-1)^{(2a-1)}(\Beta(-1; 1, 2a)+\Beta(\infty; 1, 2a))$$.--Jasper Deng (talk) 03:10, 17 June 2014 (UTC)