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

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June 16

Evaluating an Integral

How would one prove that $\underset{0}{\overset{\mathrm{\infty }}{\int }}\left[\right(1+\sqrt[a]{x}{)}^a-x]dx=\frac{1}{2}\cdot {(\genfrac{}{}{0ex}{}{-a}{a})}^{-1}$ ? — 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.

$\frac{1}{2}+a\sum _{n=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{a}{n})(\frac{1}{n+a}+\frac{1}{n-2a})=\frac{1}{2}\cdot {(\genfrac{}{}{0ex}{}{-a}{a})}^{-1}-$ 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 $\underset{0}{\overset{\mathrm{\infty }}{\int }}((1+u{)}^a-u^{\frac{1}{a}})a{u}^{a-1}du$. 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 $u^{\frac{1}{a}}$ — 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 $\underset{-1}{\overset{\mathrm{\infty }}{\int }}(v^a-(v-1{)}^a)a(v-1{)}^{a-1}dv$. Since $a$ and $(-1{)}^a$ are constant, I finally get an expression in terms of the incomplete beta function, $a(-1{)}^{(a-1)}(\mathrm{B}(-1;a+1,a)+\mathrm{B}(\mathrm{\infty };a+1,a))-a(-1{)}^{(2a-1)}(\mathrm{B}(-1;1,2a)+\mathrm{B}(\mathrm{\infty };1,2a))$.--Jasper Deng (talk) 03:10, 17 June 2014 (UTC)