Showing $\int_{-1}^{1} \frac{125}{12}\sqrt[10]{\frac{1 + x}{1 - x}} (x^2 - x) \, dx = \phi\pi$

Question

While exploring possible applications for this new trick, I stumbled upon an entire family of integrals that "always" yield $a\pi^n$, where $a$ is an algebraic number and $n$ is a natural number. The following integral captivated me greatly:

$\boxed{\int_{-1}^{1} \frac{125}{12}\sqrt[10]{\frac{1 + x}{1 - x}} (x^2 - x) \, dx = \color{red}{\phi}\color{blue}{\pi}}\tag{1}$.

The family:

$\int_{-1}^{1}(\frac{1 + x}{1 - x} )^{\frac{1}{2}}(x^2 - x) dx = 0$.

$\int_{-1}^{1} (\frac{1 + x}{1 - x})^{\frac{1}{4}} (x^2 - x) dx = \frac{\pi}{8\sqrt{2}}$.

$\int_{-1}^{1} (\frac{1 + x}{1 - x})^{\frac{1}{6}} (x^2 - x) dx = \frac{10\pi}{81}$.

$\int_{-1}^{1} \left(\frac{1 + x}{1 - x}\right)^{\frac{1}{8}} (x^2 - x) \, dx = \frac{\sqrt{3}}{100} \left(\frac{22787}{479}\right)^{\frac{1}{4}} \pi^2.$

$\int_{-1}^{1}(\frac{1 + x}{1 - x} )^{\frac{1}{10}} (x^2 - x) dx = \frac{12}{125}\phi\pi$.

$\int_{-1}^{1}(\frac{1 + x}{1 - x})^{\frac{1}{12}} (x^2 - x) dx = \frac{15455288\pi}{94257441}$.

The family seems to beg for a generalization. Is such a thing possible?

Answer 1

Let's concider $$I(\alpha)=\int_{-1}^1\left(\frac{1+x}{1-x}\right)^\alpha x(x-1)dx;\,\,-1<\alpha<2$$ Making the substitution $\frac{1+x}{1-x}=t$ $$I(\alpha)=4\int_0^\infty\frac{t^\alpha}{(1+t)^4}(1-t)dt$$ Making the substitution $x=\frac1{1+t}$ $$=4\int_0^1(1-x)^\alpha x^{2-\alpha}dx-4\int_0^1(1-x)^{\alpha+1}x^{1-\alpha}dx$$ Integrating the second term by parts $$=4\frac{1-2\alpha}{2-\alpha}\int_0^1(1-x)^\alpha x^{2-\alpha}dx=4\frac{1-2\alpha}{2-\alpha}B\big(1+\alpha;3-\alpha\big)=\frac23\frac{1-2\alpha}{2-\alpha}\Gamma(1+\alpha)\Gamma(3-\alpha)$$ Using the Euler formula for Gamma-function $$I(\alpha)=\frac23\frac{1-2\alpha}{2-\alpha}\alpha(2-\alpha)(1-\alpha)\Gamma(\alpha)\Gamma(1-\alpha)=\frac{2\pi}3\frac{\alpha(1-\alpha)(1-2\alpha)}{\sin\pi\alpha}$$ You can quickly check the answer, for example, at $\alpha=\frac16$ and get $\displaystyle\frac{10\pi}{3^4}$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{-1}^{1} \pars{1 + x \over 1 - x}^{\nu}\pars{x^{2} - x} \,\dd x} \\[5mm] = & \ \int_{-1}^{1}\pars{1 + x}^{\nu}\pars{1 - x}^{-\nu} \bracks{\pars{1 - x} - \pars{1 + x}\pars{1 - x}}\dd x \\[5mm] = & \!\!\! \int_{-1}^{1}\!\!\! \pars{1 + x}^{\nu}\,\pars{1 - x}^{1 - \nu}\,\,\dd x\! -\! \int_{-1}^{1} \pars{1 + x}^{1 + \nu}\,\pars{1 - x}^{1 - \nu}\,\,\dd x \\[5mm] = & \!\! \int_{0}^{1}\bracks{\pars{1 + x}^{\nu}\,\pars{1 - x}^{1 - \nu} + \pars{1 + x}^{1 - \nu}\,\pars{1 - x}^{\nu}}\dd x \\ & \!\!\!\!\!\!- \int_{0}^{1}\bracks{\pars{1 + x}^{1 + \nu}\,\pars{1 - x}^{1 - \nu} + \pars{1 + x}^{1 - \nu}\,\pars{1 - x}^{1 + \nu}}\dd x \\[5mm] = &\ 4\on{B}\pars{1 + \nu,2 -\nu} - 8\on{B}\pars{2 + \nu,2 - \nu}\label{1}\tag{1} \end{align} where $\ds{\on{B}}$ is the $\ds{\it Beta\ Function}$. The last result is found with identity $\ds{\bf 8.38}.6$ of G & R In addition, with Beta-recurrence and Euler Reflection Formula: $$ \left\{\begin{array}{rcl} \ds{\on{B}\pars{1 + \nu, 2 - \nu}} & \ds{=} & \ds{\on{B}\pars{1 + \nu, 1 - \nu}{1 - \nu \over 2} = \on{B}\pars{\nu, 1 - \nu}\,\nu\,{1 - \nu \over 2}} \\ & \ds{=} & \ds{{\pi \over \sin\pars{\pi\nu}}{\nu\pars{1 - \nu} \over 2}} \\[3mm] \ds{\on{B}\pars{2 + \nu,2 - \nu}} & \ds{=} & \ds{\on{B}\pars{1 + \nu,2 - \nu}{1 + \nu \over 3} = \on{B}\pars{\nu,2 - \nu}{\nu \over 2}{1 + \nu \over 3}} \\ & \ds{=} & \ds{\on{B}\pars{\nu,1 - \nu}\pars{1 - \nu}{\nu \over 2}{1 + \nu \over 3}} \\ & \ds{=} & \ds{{\pi \over \sin\pars{\pi\nu}}{\pars{1- \nu}\nu\pars{1 + \nu} \over 6}} \end{array}\right. $$ Therefore, (\ref{1}) becomes \begin{align} & \color{#44f}{\int_{-1}^{1} \pars{1 + x \over 1 - x}^{\nu}\pars{x^{2} - x} \,\dd x} = \bbx{\color{#44f}{{2\pi \over 3\sin\pars{\pi\nu}} \nu\pars{1 -\nu}\pars{1 - 2\nu}}} \\ & \end{align}