How should i substitute $x$ in this integral in order to evaluate it $${ \int_0^{\infty}((x^{\alpha}+1)^{\frac1{\alpha}}-x)\ \mathrm dx }$$
This is how i had tried it
$$\int_{0}^{1}((x^{\alpha}+1)^{\frac1{\alpha}}-x)\ \mathrm dx +\int_1^{\infty}((x^{\alpha}+1)^{\frac1{\alpha}}-x)\ \mathrm dx $$
By substituting $\frac1{x}$ in second integral
$$\int_{0}^{1}((x^{\alpha}+1)^{\frac1{\alpha}}-x)\ \mathrm dx -\int_{1}^{0} \left(\left(\frac1{x^{\alpha}}+1\right)^{\frac1{\alpha}}-\frac1{x}\right)\frac1{x^2} \mathrm dx$$
$$\int_{0}^{1}((x^{\alpha}+1)^{\frac1{\alpha}}-x)\ \mathrm dx +\int_{0}^{1} \left(\frac{(x^{\alpha}+1)^{\frac1{\alpha}}-1}{x}\right)\frac1{x^2} \mathrm dx$$
I don't know how to continue after that
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
& \bbox[5px,#ffd]{\left.\int_{0}^{\infty}\bracks{%
\pars{x^{\alpha} + 1}^{1/\alpha} - x}
\dd x\,\right\vert_{\,\alpha\ >\ 2}}
\\[5mm] = &\
\int_{0}^{\infty}x\bracks{%
\pars{1 + x^{-\alpha}}^{1/\alpha} - 1}\dd x
\\[5mm] = & \
{1 \over \alpha}\int_{0}^{\infty}t^{\color{#f00}{-2/\alpha} - 1}\,
\bracks{\pars{1 + t}^{1/\alpha} - 1}\dd t\quad
\mbox{with}\quad x = t^{-1/\alpha}
\end{align}
The last integral can be evaluated with the $\ds{Ramanujan's\ Master\ Theorem}$. Note that
$\ds{\pars{1 + t}^{1/\alpha} - 1 =
\sum_{k = 0}^{\infty}
\color{#f00}{\bracks{k > 0}\,
{\Gamma\pars{k - 1/\alpha} \over \Gamma\pars{-1/\alpha}}}
{\pars{-t}^{k} \over k!}}$. Then,
\begin{align}
& \bbox[5px,#ffd]{\left.\int_{0}^{\infty}\bracks{%
\pars{x^{\alpha} + 1}^{1/\alpha} - x}
\dd x\,\right\vert_{\,\alpha\ >\ 2}}
\\ = &\
{1 \over \alpha}\,\Gamma\pars{-\,{2 \over \alpha}}\bracks{{2 \over \alpha} > 0}
{\Gamma\pars{2/\alpha - 1/\alpha} \over \Gamma\pars{-1/\alpha}}
\\[5mm] = &\
{\Gamma\pars{-2/\alpha} \over \Gamma\pars{-1/\alpha}}
\overbrace{{1 \over \alpha}\,\Gamma\pars{1 \over \alpha}}^{\ds{\Gamma\pars{1 + {1 \over \alpha}}}}
\\[5mm] = &\
\bbx{{2^{-\pars{2 + \alpha}/\alpha}
\,\,\,\,\Gamma\pars{1/2 - 1/\alpha}\Gamma\pars{1 + 1/\alpha} \over \root{\pi}}} \\ &\
\end{align}
In the last step, I used the
Gamma Duplication Formula.
