Dtyjlui

Through some calculation, I found that for all $r>0$
$$
begin{array}{rcl}
{displaystyleint_{0}^{1}left[left(1 - x^{r}right)^{1/r} - xright]^{2},mathrm{d}x} & {displaystyle =} &
{displaystyle{1 over 3}}
\
{displaystyleint_{0}^{1}left[left(1 - x^{r}right)^{1/r} - xright]^{4},mathrm{d}x} & {displaystyle =} &
{displaystyle{1 over 5}}
\
{displaystyleint_{0}^{1}left[left(1 - x^{r}right)^{1/r} - xright]^{6},mathrm{d}x} & {displaystyle =} &
{displaystyle{1 over 7}}
end{array}
$$

It seems like for $left{n = 2k, r > 0 mid k in mathbb{N}right}$

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

I want to prove this general form.

Someone suggested to make the substitution
$$y=(1-x^r)^{1/r}$$
So 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?

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 = 2int_{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 anternative 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,

$$ I(r) - I(s)
= int_{partial mathcal{D}} ( y - x )^n , mathrm{d}x
= - iint_{mathcal{D}} n (y - x)^{n-1} , mathrm{d}xmathrm{d}y. $$

But since the region $mathcal{D}$ is symmetric around $y = x$ and $n$ is even, interchanging the roles of $x$ and $y$ shows

$$ iint_{mathcal{D}} n (y - x)^{n-1} , mathrm{d}xmathrm{d}y
= iint_{mathcal{D}} n (x - y)^{n-1} , mathrm{d}xmathrm{d}y
= - iint_{mathcal{D}} n (y - x)^{n-1} , mathrm{d}xmathrm{d}y. $$

Therefore $I(r) = I(s)$ for any $ r < s$, and in particular, letting $s to infty$ gives

$$ I(r) = int _{0}^{1} (1 - x)^n , mathrm{d}x = frac{1}{n+1}. $$

Here's your Beta integral
$$S=int_0^1x^{n-k}(1-x^r)^{k/r}dx$$
Setting $w=x^r$, we see that
$$S=frac1rint_0^1w^{frac{n+1-k-r}r}(1-w)^{k/r}dw$$
$$S=frac1rint_0^1w^{frac{n+1-k}r-1}(1-w)^{frac{k+r}r-1}dw$$
$$S=frac1rmathrm{B}bigg(frac{n+1-k}r,frac{k+r}rbigg)$$
So
$$I(r,n)=int_0^1[(1-x^r)^{1/r}-x]^ndx$$
$$I(r,n)=frac1rsum_{k=0}^{n}(-1)^{n-k}{nchoose k}frac{Gamma(frac{n+1-k}r)Gamma(frac{k+r}r)}{Gamma(1+frac{n+1}r)}$$
$$I(r,n)=frac1{r}frac{Gamma(n+1)}{Gamma(1+frac{n+1}r)}sum_{k=0}^{n}(-1)^{n-k}frac{Gamma(frac{n+1-k}r)Gamma(frac{k+r}r)}{Gamma(k+1)Gamma(n-k+1)}$$

Which is a closed form

Here's a proof with Hypergeometirc function.

We have
$$
underset{j=1}{overset{2 n+1}{sum }}
left(
begin{array}{c}
2 n \
j-1 \
end{array}
right)
(-x)^{j-1}
left(left(1-x^rright)^{1/r}right)^{-j+2 n+1}
=left(left(1-x^rright)^{1/r}-xright)^{2 n}
$$
by binomial expansion.

It is easy to verify that
$$
left(
begin{array}{c}
2 n \
j-1 \
end{array}
right)
(-x)^{j-1}
left(left(1-x^rright)^{1/r}right)^{-j+2 n+1}
=
frac{mathrm d}{mathrm d x}left(
frac{1}{2 n+1}
(-1)^{j+1} x^j binom{2 n+1}{j} , _2F_1left(frac{j}{r},-frac{-j+2 n+1}{r};frac{j}{r}+1;x^rright)
right)
$$
Therefore, we have
$$
int((1-x^r)^{1/r}-x)^{2 n} mathrm dx
=
sum _{j=1}^{2 n+1} frac{1}{2 n+1} (-1)^{j+1} x^j binom{2 n+1}{j} , _2F_1left(frac{j}{r},-frac{-j+2 n+1}{r};frac{j}{r}+1;x^rright).
$$
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 $frac 1 {2n+1}$.