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

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June 16

Evaluating an Integral

How would one prove that $\int _{0}^{\infty }{\bigg [}{\Big (}1+{\sqrt[{a}]{x}}{\Big )}^{a}-x{\bigg ]}dx~=~{\frac {1}{2}}\cdot {-a \choose a}^{-1}$ ? — 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.

${\frac {1}{2}}+a\sum _{n=1}^{\infty }{a \choose n}{\Bigg (}{\frac {1}{n+a}}+{\frac {1}{n-2a}}{\Bigg )}={\frac {1}{2}}\cdot {-a \choose a}^{-1}~-$ 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 $u^{\frac {1}{a}}\quad$ — 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, $a(-1)^{(a-1)}(\mathrm {B} (-1;a+1,a)+\mathrm {B} (\infty ;a+1,a))-a(-1)^{(2a-1)}(\mathrm {B} (-1;1,2a)+\mathrm {B} (\infty ;1,2a))$.--Jasper Deng (talk) 03:10, 17 June 2014 (UTC)