Proving that $\int_0^\infty\Big(\sqrt[n]{1+x^n}-x\Big)dx~=~\frac12\cdot{-1/n\choose+1/n}^{-1}$

Solution 1:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x =\half\,{-1/n \choose 1/n}^{-1}:\ {\large ?}.\qquad\qquad n > 2}$.

\begin{align}&\overbrace{\color{#c00000}{ \int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x}} ^{\ds{x^{n}\ \mapsto x}}\ =\ \int_{0}^{\infty}\bracks{\pars{1 + x}^{1/n} - x^{1/n}}\,{1 \over n}\,x^{1/n - 1} \,\dd x \\[3mm]&={1 \over n}\ \overbrace{\int_{0}^{\infty}\bracks{ \pars{1 + x}^{1/n}x^{1/n - 1} - x^{2/n - 1}}\,\dd x} ^{\ds{t\ \equiv {1 \over 1 + x}\ \imp\ x = {1 \over t} - 1}} \\[3mm]&={1 \over n}\int_{1}^{0}\bracks{ t^{-1/n}\pars{{1 \over t} - 1}^{1/n - 1} -\pars{{1 \over t} - 1}^{2/n - 1}}\,\pars{-\,{\dd t \over t^{2}}} \\[3mm]&={1 \over n}\int_{0}^{1}\bracks{ t^{-2/n - 1}\pars{1 - t}^{1/n - 1} - t^{-2/n - 1}\pars{1 - t}^{2/n - 1}}\,\dd t \end{align}

\begin{align}&\color{#c00000}{ \int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x} ={1 \over n}\int_{0}^{1}{t^{1/n - 1} - 1 \over \pars{1 - t}^{2/n + 1}}\,\dd t- {1 \over n}\int_{0}^{1}{t^{2/n - 1} - 1 \over \pars{1 - t}^{2/n + 1}}\,\dd t \end{align} Both integrals converge when $\ds{\Re\pars{n} > 2}$. Moreover, $$ \int_{0}^{1}{t^{a} - 1 \over \pars{1 - t}^{2/n + 1}}\,\dd t =\half\,n + {a!\pars{-2/n - 1}! \over \pars{a - 2/n}!} $$

such that \begin{align}&\left.\color{#66f}{\large \int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x}\,\right\vert_{\ n\ >\ 2} ={1 \over n}\,{\pars{1/n - 1}!\pars{-2/n - 1}! \over \pars{-1/n - 1}!} \\[3mm]&={1 \over n}\, {\bracks{n\pars{1/n}!}\bracks{\pars{-n/2}\pars{-2/n}!} \over -n\pars{-1/n}!} =\half\bracks{\pars{-1/n}! \over \pars{1/n}!\pars{-2/n}!}^{-1} =\color{#66f}{\large\half\,{-1/n \choose 1/n}^{-1}} \end{align}