Ways to show that $\int_{0}^{1}((1-x^r)^{1/r}-x)^{2k}dx=\frac{1}{2k+1}$ Through some calculation, I found that for all $r>0$
$$
\begin{array}{rcl}
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{2}\,\mathrm{d}x} &  {\displaystyle =} &
{\displaystyle{1 \over 3}}
\\
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{4}\,\mathrm{d}x} &  {\displaystyle =} &
{\displaystyle{1 \over 5}}
\\
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{6}\,\mathrm{d}x} &  {\displaystyle =} &
{\displaystyle{1 \over 7}}
\end{array}
$$
It seems like for  $k \in \mathbb{N}$ 

$$
\int_{0}^{1}
\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{2k}\mathrm{d}x = {1 \over 2k + 1}
$$

I want to prove this general form.
Someone suggested to make the substitution 
$$y=(1-x^r)^{1/r}$$
Let $n=2k$, and I rewrote the integral into 
$$\int_{0}^{1}((1-x^r)^{1/r}-x)^ndx=\int_{0}^{1}(y-x)^ndx$$
and tried to use the binomial formula:
$$(y-x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}x^{n-k}y^{k}$$
The integral then becomes
$$\begin{align}
\int_{0}^{1}(y-x)^ndx&=\int_{0}^{1}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}x^{n-k}y^{k}dx\\
&=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}\int_{0}^{1}x^{n-k}y^{k}dx\\
&=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}\int_{0}^{1}x^{n-k}(1-x^r)^{k/r}dx\\
\end{align}$$
Now I think I need to use Beta function:
$$B(x,y) = \frac{(x-1)!(y-1)!}{(x+y-1)!}= \int_{0}^{1}u^{x-1}(1-u)^{y-1}du=\sum_{n=0}^{\infty}\frac{{\binom{n-y}{n}}}{x+n}$$
Am I on the right track? Are here any easier ways to prove the general form?
 A: Here's your Beta integral
$$S=\int_0^1x^{n-k}(1-x^r)^{k/r}dx$$
Setting $w=x^r$, we see that
$$S=\frac1r\int_0^1w^{\frac{n+1-k-r}r}(1-w)^{k/r}dw$$
$$S=\frac1r\int_0^1w^{\frac{n+1-k}r-1}(1-w)^{\frac{k+r}r-1}dw$$
$$S=\frac1r\mathrm{B}\bigg(\frac{n+1-k}r,\frac{k+r}r\bigg)$$
So 
$$I(r,n)=\int_0^1[(1-x^r)^{1/r}-x]^ndx$$
$$I(r,n)=\frac1r\sum_{k=0}^{n}(-1)^{n-k}{n\choose k}\frac{\Gamma(\frac{n+1-k}r)\Gamma(\frac{k+r}r)}{\Gamma(1+\frac{n+1}r)}$$
$$I(r,n)=\frac1{r}\frac{\Gamma(n+1)}{\Gamma(1+\frac{n+1}r)}\sum_{k=0}^{n}(-1)^{n-k}\frac{\Gamma(\frac{n+1-k}r)\Gamma(\frac{k+r}r)}{\Gamma(k+1)\Gamma(n-k+1)}$$
Which is a closed form
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{1}\bracks{\pars{1 - x^{r}}^{1/r} - x}^{2k}\,\dd x\,\right\vert_{{\large r\ >\ 0} \atop
{\large k\ \in\ \mathbb{N}_{\geq\ 0}}}}
\\[5mm] \stackrel{x^{\large r}\ \mapsto\ x}{=}\,\,\,&
\int_{0}^{1}\bracks{\pars{1 - x}^{1/r} - x^{1/r}}^{2k}\,{1 \over r}\,
x^{1/r - 1}\,\dd x
\\[5mm]
\,\,\,\stackrel{x\ \mapsto\ x + 1/2}{=}\,\,\,&
{1 \over r}\int_{-1/2}^{1/2}\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x}^{1/r}}^{2k}\,
\pars{{1 \over 2} + x}^{1/r - 1}\,\dd x
\\[8mm] = &\
{1 \over r}\int_{0}^{1/2}\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x\!}^{1/r}}^{2k}\times
\\[2mm] &\
\phantom{{1 \over r}\int_{0}^{1/2}}\bracks{\pars{{1 \over 2} + x}^{1/r - 1} +
\pars{{1 \over 2} - x}^{1/r - 1}\!}\!\dd x
\\[8mm] = &\
-\int_{0}^{1/2}{1 \over 2k + 1}\,\partiald{}{x}\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x}^{1/r}}^{2k + 1}\,\dd x
\\[5mm] = &\
\underbrace{\braces{-\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x}^{1/r}}^{2k + 1}}_{x\ =\ 0}^{x\ =\ 1/2}}
_{\ds{=\ 1\ -\ 0\ =\ 1}}\,\,\,{1 \over 2k + 1}
\\[5mm] = &\
\bbx{1 \over 2k + 1}
\end{align}
A: Here's a proof with Hypergeometirc function.
We have
$$
\underset{j=1}{\overset{2 n+1}{\sum }} 
\left(
\begin{array}{c}
 2 n \\
 j-1 \\
\end{array}
\right) 
(-x)^{j-1}
\left(\left(1-x^r\right)^{1/r}\right)^{-j+2 n+1}
=\left(\left(1-x^r\right)^{1/r}-x\right)^{2 n}
$$
by binomial expansion.
It is easy to verify that
$$
\left(
\begin{array}{c}
 2 n \\
 j-1 \\
\end{array}
\right) 
(-x)^{j-1}
\left(\left(1-x^r\right)^{1/r}\right)^{-j+2 n+1}
=
\frac{\mathrm d}{\mathrm d x}\left(
\frac{1}{2 n+1}
(-1)^{j+1} x^j \binom{2 n+1}{j} \, _2F_1\left(\frac{j}{r},-\frac{-j+2 n+1}{r};\frac{j}{r}+1;x^r\right)
\right)
$$
Therefore, we have
$$
\int((1-x^r)^{1/r}-x)^{2 n} \mathrm dx
=
\sum _{j=1}^{2 n+1} \frac{1}{2 n+1} (-1)^{j+1} x^j \binom{2 n+1}{j} \, _2F_1\left(\frac{j}{r},-\frac{-j+2 n+1}{r};\frac{j}{r}+1;x^r\right).
$$
When $j=2n+1$, the summand in the right hand equals $\frac{x^{2 n+1}}{2 n+1}$. This is the term which gives us $\frac 1 {2n+1}$.
A: We present 3 different solutions.

Solution 1 - slick substitution. We prove a more general statement:

Proposition. Let $R \in (0, \infty]$ and let $\varphi : [0, R] \to [0, R]$ satisfy the following conditions:

*

*$\varphi$ is continuous on $[0, R]$;

*$\varphi(0) = R$ and $\varphi(R) = 0$;

*$\varphi$ is bijective and $\varphi^{-1} = \varphi$.

Then for any integrable function $f$ on $[0, R]$,
$$ \int_{0}^{R} f(|x-\varphi(x)|) \, \mathrm{d}x = \int_{0}^{R} f(x) \, \mathrm{d}x. $$

Proof. In case $\varphi$ is also continuously differentiable on $(0, R)$, by the substitution $y = \varphi(x)$, or equivalently, $x = \varphi(y)$,
$$ I
:= \int_{0}^{R} f( |x - \varphi(x)| ) \, \mathrm{d}x
= -\int_{0}^{R} f( |\varphi(y) - y| ) \varphi'(y) \, \mathrm{d}y. $$
Summing two integrals,
\begin{align*}
2I
&= \int_{0}^{R} f( |x - \varphi(x)| ) (1 - \varphi'(x)) \, \mathrm{d}x \\
&= \int_{-R}^{R} f( |u| ) \, \mathrm{d}u = 2\int_{0}^{R} f(u) \, \mathrm{d}u, \tag{$u = x - \varphi(x)$}
\end{align*}
proving the claim when $\varphi$ is continuously differentiable. This proof can be easily adapted to general $\varphi$ by using Stieltjes integral. ■
Now plug $\varphi(x) = (1-x^r)^{1/r}$ with $R = 1$ and $f(x) = x^n$ for positive even integer $n$. Then
$$ \int_{0}^{1} \left( (1-x^r)^{1/r} - x \right)^n \, \mathrm{d}x
= \int_{0}^{1} \left| x - (1-x^r)^{1/r} \right|^n \, \mathrm{d}x
= \int_{0}^{1} x^n \, \mathrm{d}x
= \frac{1}{n+1}. $$

Solution 2 - using beta function. Here is an alternative solution. Write $p = 1/r$. Then using the substitution $x = u^p$,
\begin{align*}
\int_{0}^{1} \left( (1 - x^r)^{1/r} - x \right)^n \, \mathrm{d}x
&= \int_{0}^{1} \left( (1 - u)^{p} - u^p \right)^{n} pu^{p-1} \, \mathrm{d}u \\
&= \sum_{k=0}^{n} (-1)^k \binom{n}{k} p \int_{0}^{1} (1-u)^{p(n-k)} u^{p(k+1)-1} \, \mathrm{d}u \\
&= \sum_{k=0}^{n} (-1)^k \binom{n}{k} p \cdot \frac{(p(n-k))!(p(k+1)-1)!}{(p(n+1))!}
\end{align*}
Here, $s! = \Gamma(s+1)$. Now define $a_k = (pk)!/k!$. Then the above sum simplifies to
\begin{align*}
\int_{0}^{1} \left( (1 - x^r)^{1/r} - x \right)^n \, \mathrm{d}x
&= \frac{1}{(n+1)a_{n+1}} \sum_{k=0}^{n} (-1)^k a_{n-k}a_{k+1} \\
&= \frac{1}{(n+1)a_{n+1}} \left( a_0 a_{n+1} + \sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} \right).
\end{align*}
So it suffices to show that $\sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} = 0$. But by the substitution $l = n-1-k$, we have
$$ \sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1}
= - \sum_{l=0}^{n-1} (-1)^l a_{l+1}a_{n-l}. $$
(Here the parity of $n$ is used.) So the sum equals its negation, hence is zero as required.

Solution 3 - using multivariate calculus. Let $\mathcal{C}_r$ denote the curve defined by $x^r + y^r = 1$ in the first quadrant, oriented to the right. Then
$$ I(r) := \int_{0}^{1} \left( (1 -x^r)^{1/r} - x \right)^n \, \mathrm{d}x
= \int_{\mathcal{C}_r} ( y - x )^n \, \mathrm{d}x. $$
Notice that if $0 < r < s$, then $\mathcal{C}_s$ lies above $\mathcal{C}_r$, and so, the curve $\mathcal{C}_r - \mathcal{C}_s$ bounds some region, which we denote by $\mathcal{D}$, counter-clockwise:
$\hspace{10em}$ 
Then by Green's theorem,
$$ I(r) - I(s)
= \int_{\partial \mathcal{D}} ( y - x )^n \, \mathrm{d}x
= - \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y. $$
But since the region $\mathcal{D}$ is symmetric around $y = x$ and $n$ is even, interchanging the roles of $x$ and $y$ shows
$$ \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y
= \iint_{\mathcal{D}} n (x - y)^{n-1} \, \mathrm{d}x\mathrm{d}y
= - \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y. $$
Therefore $I(r) = I(s)$ for any $ r < s$, and in particular, letting $s \to \infty$ gives
$$ I(r) = \int _{0}^{1} (1 - x)^n \, \mathrm{d}x = \frac{1}{n+1}. $$
