Ways to show that $int_{0}^{1}((1-x^r)^{1/r}-x)^ndx=frac{1}{n+1}$












23












$begingroup$


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?










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$endgroup$












  • $begingroup$
    Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $frac{Gamma left(1+frac{1}{n}right)^2}{Gamma left(frac{n+2}{n}right)}-frac{1}{2}$
    $endgroup$
    – David G. Stork
    Jan 11 at 23:45










  • $begingroup$
    Note: switching between $n$ and $r$ between the header and the body is very confusing. Sticking with the header notation, and letting, $n=1$ you get the integral $int_0^1 (1-2x)^r,dx$ which is not always $frac 1{r+1}$. If $r=1$, say, you get $0$.
    $endgroup$
    – lulu
    Jan 12 at 0:03










  • $begingroup$
    And, sticking with the header notation and letting $n=2$, we see that $int_0^1 ((1-x^2)^{1/2}-x)^3,dx = frac {3pi}8 -1 approx .1781 neq frac 14$ . Or am I misreading something?
    $endgroup$
    – lulu
    Jan 12 at 0:08






  • 2




    $begingroup$
    I see, thank you for re-asking the question! It is a nice integral for sure.
    $endgroup$
    – Zacky
    Jan 12 at 0:26






  • 1




    $begingroup$
    @clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out.
    $endgroup$
    – Larry
    Jan 12 at 0:51


















23












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $frac{Gamma left(1+frac{1}{n}right)^2}{Gamma left(frac{n+2}{n}right)}-frac{1}{2}$
    $endgroup$
    – David G. Stork
    Jan 11 at 23:45










  • $begingroup$
    Note: switching between $n$ and $r$ between the header and the body is very confusing. Sticking with the header notation, and letting, $n=1$ you get the integral $int_0^1 (1-2x)^r,dx$ which is not always $frac 1{r+1}$. If $r=1$, say, you get $0$.
    $endgroup$
    – lulu
    Jan 12 at 0:03










  • $begingroup$
    And, sticking with the header notation and letting $n=2$, we see that $int_0^1 ((1-x^2)^{1/2}-x)^3,dx = frac {3pi}8 -1 approx .1781 neq frac 14$ . Or am I misreading something?
    $endgroup$
    – lulu
    Jan 12 at 0:08






  • 2




    $begingroup$
    I see, thank you for re-asking the question! It is a nice integral for sure.
    $endgroup$
    – Zacky
    Jan 12 at 0:26






  • 1




    $begingroup$
    @clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out.
    $endgroup$
    – Larry
    Jan 12 at 0:51
















23












23








23


20



$begingroup$


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?










share|cite|improve this question











$endgroup$




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?







calculus integration definite-integrals






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share|cite|improve this question













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share|cite|improve this question








edited Mar 5 at 0:48









Felix Marin

68.4k7109144




68.4k7109144










asked Jan 11 at 23:26









LarryLarry

2,52831131




2,52831131












  • $begingroup$
    Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $frac{Gamma left(1+frac{1}{n}right)^2}{Gamma left(frac{n+2}{n}right)}-frac{1}{2}$
    $endgroup$
    – David G. Stork
    Jan 11 at 23:45










  • $begingroup$
    Note: switching between $n$ and $r$ between the header and the body is very confusing. Sticking with the header notation, and letting, $n=1$ you get the integral $int_0^1 (1-2x)^r,dx$ which is not always $frac 1{r+1}$. If $r=1$, say, you get $0$.
    $endgroup$
    – lulu
    Jan 12 at 0:03










  • $begingroup$
    And, sticking with the header notation and letting $n=2$, we see that $int_0^1 ((1-x^2)^{1/2}-x)^3,dx = frac {3pi}8 -1 approx .1781 neq frac 14$ . Or am I misreading something?
    $endgroup$
    – lulu
    Jan 12 at 0:08






  • 2




    $begingroup$
    I see, thank you for re-asking the question! It is a nice integral for sure.
    $endgroup$
    – Zacky
    Jan 12 at 0:26






  • 1




    $begingroup$
    @clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out.
    $endgroup$
    – Larry
    Jan 12 at 0:51




















  • $begingroup$
    Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $frac{Gamma left(1+frac{1}{n}right)^2}{Gamma left(frac{n+2}{n}right)}-frac{1}{2}$
    $endgroup$
    – David G. Stork
    Jan 11 at 23:45










  • $begingroup$
    Note: switching between $n$ and $r$ between the header and the body is very confusing. Sticking with the header notation, and letting, $n=1$ you get the integral $int_0^1 (1-2x)^r,dx$ which is not always $frac 1{r+1}$. If $r=1$, say, you get $0$.
    $endgroup$
    – lulu
    Jan 12 at 0:03










  • $begingroup$
    And, sticking with the header notation and letting $n=2$, we see that $int_0^1 ((1-x^2)^{1/2}-x)^3,dx = frac {3pi}8 -1 approx .1781 neq frac 14$ . Or am I misreading something?
    $endgroup$
    – lulu
    Jan 12 at 0:08






  • 2




    $begingroup$
    I see, thank you for re-asking the question! It is a nice integral for sure.
    $endgroup$
    – Zacky
    Jan 12 at 0:26






  • 1




    $begingroup$
    @clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out.
    $endgroup$
    – Larry
    Jan 12 at 0:51


















$begingroup$
Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $frac{Gamma left(1+frac{1}{n}right)^2}{Gamma left(frac{n+2}{n}right)}-frac{1}{2}$
$endgroup$
– David G. Stork
Jan 11 at 23:45




$begingroup$
Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $frac{Gamma left(1+frac{1}{n}right)^2}{Gamma left(frac{n+2}{n}right)}-frac{1}{2}$
$endgroup$
– David G. Stork
Jan 11 at 23:45












$begingroup$
Note: switching between $n$ and $r$ between the header and the body is very confusing. Sticking with the header notation, and letting, $n=1$ you get the integral $int_0^1 (1-2x)^r,dx$ which is not always $frac 1{r+1}$. If $r=1$, say, you get $0$.
$endgroup$
– lulu
Jan 12 at 0:03




$begingroup$
Note: switching between $n$ and $r$ between the header and the body is very confusing. Sticking with the header notation, and letting, $n=1$ you get the integral $int_0^1 (1-2x)^r,dx$ which is not always $frac 1{r+1}$. If $r=1$, say, you get $0$.
$endgroup$
– lulu
Jan 12 at 0:03












$begingroup$
And, sticking with the header notation and letting $n=2$, we see that $int_0^1 ((1-x^2)^{1/2}-x)^3,dx = frac {3pi}8 -1 approx .1781 neq frac 14$ . Or am I misreading something?
$endgroup$
– lulu
Jan 12 at 0:08




$begingroup$
And, sticking with the header notation and letting $n=2$, we see that $int_0^1 ((1-x^2)^{1/2}-x)^3,dx = frac {3pi}8 -1 approx .1781 neq frac 14$ . Or am I misreading something?
$endgroup$
– lulu
Jan 12 at 0:08




2




2




$begingroup$
I see, thank you for re-asking the question! It is a nice integral for sure.
$endgroup$
– Zacky
Jan 12 at 0:26




$begingroup$
I see, thank you for re-asking the question! It is a nice integral for sure.
$endgroup$
– Zacky
Jan 12 at 0:26




1




1




$begingroup$
@clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out.
$endgroup$
– Larry
Jan 12 at 0:51






$begingroup$
@clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out.
$endgroup$
– Larry
Jan 12 at 0:51












4 Answers
4






active

oldest

votes


















30





+50







$begingroup$

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:



$hspace{10em}$ Region D and two enclosing curves



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



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






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Beautiful! This looks like Glasser's Master theorem little brother :D
    $endgroup$
    – Zacky
    Jan 12 at 1:02












  • $begingroup$
    Was it necessary to start with absolute value in the argument of function?
    $endgroup$
    – user
    Jan 12 at 1:04










  • $begingroup$
    @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
    $endgroup$
    – Sangchul Lee
    Jan 12 at 1:06








  • 1




    $begingroup$
    @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
    $endgroup$
    – Sangchul Lee
    Jan 12 at 1:12












  • $begingroup$
    @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
    $endgroup$
    – Larry
    Jan 27 at 23:17



















5












$begingroup$

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






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  • 1




    $begingroup$
    Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
    $endgroup$
    – user
    Jan 12 at 1:45












  • $begingroup$
    @user Ah yes I forgot $n$ was even
    $endgroup$
    – clathratus
    Jan 12 at 1:57






  • 1




    $begingroup$
    Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
    $endgroup$
    – DavidG
    Jan 12 at 5:31



















4












$begingroup$

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 $frac 1 {2n+1}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$

    begin{align}
    &bbox[10px,#ffd]{left.int_{0}^{1}bracks{pars{1 - x^{r}}^{1/r} - x}^{2k},dd x,rightvert_{{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
    \[5mm] = &
    {1 over r}int_{0}^{1/2}!!bracks{!!pars{{1 over 2} - x}^{1/r} - pars{{1 over 2} + x!}^{1/r}}^{2k}!
    bracks{!pars{{1 over 2} + x}^{1/r - 1} +
    pars{{1 over 2} - x}^{1/r - 1}!}!dd x
    \[5mm] = &
    -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} =
    bbx{1 over 2k + 1}
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice solution, (+1).
      $endgroup$
      – Larry
      Mar 5 at 13:19










    • $begingroup$
      Thanks @Larry .
      $endgroup$
      – Felix Marin
      Mar 5 at 15:23











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    4 Answers
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    4 Answers
    4






    active

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    active

    oldest

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    30





    +50







    $begingroup$

    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:



    $hspace{10em}$ Region D and two enclosing curves



    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



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






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Beautiful! This looks like Glasser's Master theorem little brother :D
      $endgroup$
      – Zacky
      Jan 12 at 1:02












    • $begingroup$
      Was it necessary to start with absolute value in the argument of function?
      $endgroup$
      – user
      Jan 12 at 1:04










    • $begingroup$
      @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:06








    • 1




      $begingroup$
      @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:12












    • $begingroup$
      @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
      $endgroup$
      – Larry
      Jan 27 at 23:17
















    30





    +50







    $begingroup$

    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:



    $hspace{10em}$ Region D and two enclosing curves



    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



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






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Beautiful! This looks like Glasser's Master theorem little brother :D
      $endgroup$
      – Zacky
      Jan 12 at 1:02












    • $begingroup$
      Was it necessary to start with absolute value in the argument of function?
      $endgroup$
      – user
      Jan 12 at 1:04










    • $begingroup$
      @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:06








    • 1




      $begingroup$
      @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:12












    • $begingroup$
      @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
      $endgroup$
      – Larry
      Jan 27 at 23:17














    30





    +50







    30





    +50



    30




    +50



    $begingroup$

    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:



    $hspace{10em}$ Region D and two enclosing curves



    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



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






    share|cite|improve this answer











    $endgroup$



    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:



    $hspace{10em}$ Region D and two enclosing curves



    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



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







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 28 at 0:56

























    answered Jan 12 at 0:53









    Sangchul LeeSangchul Lee

    95.6k12171279




    95.6k12171279








    • 1




      $begingroup$
      Beautiful! This looks like Glasser's Master theorem little brother :D
      $endgroup$
      – Zacky
      Jan 12 at 1:02












    • $begingroup$
      Was it necessary to start with absolute value in the argument of function?
      $endgroup$
      – user
      Jan 12 at 1:04










    • $begingroup$
      @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:06








    • 1




      $begingroup$
      @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:12












    • $begingroup$
      @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
      $endgroup$
      – Larry
      Jan 27 at 23:17














    • 1




      $begingroup$
      Beautiful! This looks like Glasser's Master theorem little brother :D
      $endgroup$
      – Zacky
      Jan 12 at 1:02












    • $begingroup$
      Was it necessary to start with absolute value in the argument of function?
      $endgroup$
      – user
      Jan 12 at 1:04










    • $begingroup$
      @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:06








    • 1




      $begingroup$
      @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
      $endgroup$
      – Sangchul Lee
      Jan 12 at 1:12












    • $begingroup$
      @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
      $endgroup$
      – Larry
      Jan 27 at 23:17








    1




    1




    $begingroup$
    Beautiful! This looks like Glasser's Master theorem little brother :D
    $endgroup$
    – Zacky
    Jan 12 at 1:02






    $begingroup$
    Beautiful! This looks like Glasser's Master theorem little brother :D
    $endgroup$
    – Zacky
    Jan 12 at 1:02














    $begingroup$
    Was it necessary to start with absolute value in the argument of function?
    $endgroup$
    – user
    Jan 12 at 1:04




    $begingroup$
    Was it necessary to start with absolute value in the argument of function?
    $endgroup$
    – user
    Jan 12 at 1:04












    $begingroup$
    @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
    $endgroup$
    – Sangchul Lee
    Jan 12 at 1:06






    $begingroup$
    @user, It is kind of necessary, in the sense that $$ int_{0}^{1} gleft(x-(1-x^r)^{1/r}right) , mathrm{d}x = int_{0}^{1} g(u) , mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.
    $endgroup$
    – Sangchul Lee
    Jan 12 at 1:06






    1




    1




    $begingroup$
    @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
    $endgroup$
    – Sangchul Lee
    Jan 12 at 1:12






    $begingroup$
    @Zacky, Both Glasser's master theorem and my answer deals with specific examples of measure-preserving transformations, hence the conclusion should look similar. Of course, the beauty of Glasser's result is that its proof is very elementary. (The result itself was known much prior to his paper.)
    $endgroup$
    – Sangchul Lee
    Jan 12 at 1:12














    $begingroup$
    @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
    $endgroup$
    – Larry
    Jan 27 at 23:17




    $begingroup$
    @SangchulLee: You mentioned to use the fact that $|x - varphi(x)|^n = (varphi(x) - x)^n$. Is there a name for this fact. I don't quite understand how you get there.
    $endgroup$
    – Larry
    Jan 27 at 23:17











    5












    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
      $endgroup$
      – user
      Jan 12 at 1:45












    • $begingroup$
      @user Ah yes I forgot $n$ was even
      $endgroup$
      – clathratus
      Jan 12 at 1:57






    • 1




      $begingroup$
      Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
      $endgroup$
      – DavidG
      Jan 12 at 5:31
















    5












    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
      $endgroup$
      – user
      Jan 12 at 1:45












    • $begingroup$
      @user Ah yes I forgot $n$ was even
      $endgroup$
      – clathratus
      Jan 12 at 1:57






    • 1




      $begingroup$
      Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
      $endgroup$
      – DavidG
      Jan 12 at 5:31














    5












    5








    5





    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$



    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







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 12 at 1:56

























    answered Jan 12 at 0:59









    clathratusclathratus

    4,8901338




    4,8901338








    • 1




      $begingroup$
      Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
      $endgroup$
      – user
      Jan 12 at 1:45












    • $begingroup$
      @user Ah yes I forgot $n$ was even
      $endgroup$
      – clathratus
      Jan 12 at 1:57






    • 1




      $begingroup$
      Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
      $endgroup$
      – DavidG
      Jan 12 at 5:31














    • 1




      $begingroup$
      Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
      $endgroup$
      – user
      Jan 12 at 1:45












    • $begingroup$
      @user Ah yes I forgot $n$ was even
      $endgroup$
      – clathratus
      Jan 12 at 1:57






    • 1




      $begingroup$
      Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
      $endgroup$
      – DavidG
      Jan 12 at 5:31








    1




    1




    $begingroup$
    Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
    $endgroup$
    – user
    Jan 12 at 1:45






    $begingroup$
    Try to find an error in your derivation as $I(r,n)=frac{1}{n+1}$ for even $n$ and any $r$.
    $endgroup$
    – user
    Jan 12 at 1:45














    $begingroup$
    @user Ah yes I forgot $n$ was even
    $endgroup$
    – clathratus
    Jan 12 at 1:57




    $begingroup$
    @user Ah yes I forgot $n$ was even
    $endgroup$
    – clathratus
    Jan 12 at 1:57




    1




    1




    $begingroup$
    Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
    $endgroup$
    – DavidG
    Jan 12 at 5:31




    $begingroup$
    Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)
    $endgroup$
    – DavidG
    Jan 12 at 5:31











    4












    $begingroup$

    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 $frac 1 {2n+1}$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      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 $frac 1 {2n+1}$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        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 $frac 1 {2n+1}$.






        share|cite|improve this answer









        $endgroup$



        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 $frac 1 {2n+1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 11:03









        ablmfablmf

        2,56842452




        2,56842452























            1












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$

            begin{align}
            &bbox[10px,#ffd]{left.int_{0}^{1}bracks{pars{1 - x^{r}}^{1/r} - x}^{2k},dd x,rightvert_{{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
            \[5mm] = &
            {1 over r}int_{0}^{1/2}!!bracks{!!pars{{1 over 2} - x}^{1/r} - pars{{1 over 2} + x!}^{1/r}}^{2k}!
            bracks{!pars{{1 over 2} + x}^{1/r - 1} +
            pars{{1 over 2} - x}^{1/r - 1}!}!dd x
            \[5mm] = &
            -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} =
            bbx{1 over 2k + 1}
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice solution, (+1).
              $endgroup$
              – Larry
              Mar 5 at 13:19










            • $begingroup$
              Thanks @Larry .
              $endgroup$
              – Felix Marin
              Mar 5 at 15:23
















            1












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$

            begin{align}
            &bbox[10px,#ffd]{left.int_{0}^{1}bracks{pars{1 - x^{r}}^{1/r} - x}^{2k},dd x,rightvert_{{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
            \[5mm] = &
            {1 over r}int_{0}^{1/2}!!bracks{!!pars{{1 over 2} - x}^{1/r} - pars{{1 over 2} + x!}^{1/r}}^{2k}!
            bracks{!pars{{1 over 2} + x}^{1/r - 1} +
            pars{{1 over 2} - x}^{1/r - 1}!}!dd x
            \[5mm] = &
            -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} =
            bbx{1 over 2k + 1}
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice solution, (+1).
              $endgroup$
              – Larry
              Mar 5 at 13:19










            • $begingroup$
              Thanks @Larry .
              $endgroup$
              – Felix Marin
              Mar 5 at 15:23














            1












            1








            1





            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$

            begin{align}
            &bbox[10px,#ffd]{left.int_{0}^{1}bracks{pars{1 - x^{r}}^{1/r} - x}^{2k},dd x,rightvert_{{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
            \[5mm] = &
            {1 over r}int_{0}^{1/2}!!bracks{!!pars{{1 over 2} - x}^{1/r} - pars{{1 over 2} + x!}^{1/r}}^{2k}!
            bracks{!pars{{1 over 2} + x}^{1/r - 1} +
            pars{{1 over 2} - x}^{1/r - 1}!}!dd x
            \[5mm] = &
            -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} =
            bbx{1 over 2k + 1}
            end{align}






            share|cite|improve this answer











            $endgroup$



            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$

            begin{align}
            &bbox[10px,#ffd]{left.int_{0}^{1}bracks{pars{1 - x^{r}}^{1/r} - x}^{2k},dd x,rightvert_{{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
            \[5mm] = &
            {1 over r}int_{0}^{1/2}!!bracks{!!pars{{1 over 2} - x}^{1/r} - pars{{1 over 2} + x!}^{1/r}}^{2k}!
            bracks{!pars{{1 over 2} + x}^{1/r - 1} +
            pars{{1 over 2} - x}^{1/r - 1}!}!dd x
            \[5mm] = &
            -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} =
            bbx{1 over 2k + 1}
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 5 at 1:16

























            answered Mar 5 at 1:10









            Felix MarinFelix Marin

            68.4k7109144




            68.4k7109144












            • $begingroup$
              Nice solution, (+1).
              $endgroup$
              – Larry
              Mar 5 at 13:19










            • $begingroup$
              Thanks @Larry .
              $endgroup$
              – Felix Marin
              Mar 5 at 15:23


















            • $begingroup$
              Nice solution, (+1).
              $endgroup$
              – Larry
              Mar 5 at 13:19










            • $begingroup$
              Thanks @Larry .
              $endgroup$
              – Felix Marin
              Mar 5 at 15:23
















            $begingroup$
            Nice solution, (+1).
            $endgroup$
            – Larry
            Mar 5 at 13:19




            $begingroup$
            Nice solution, (+1).
            $endgroup$
            – Larry
            Mar 5 at 13:19












            $begingroup$
            Thanks @Larry .
            $endgroup$
            – Felix Marin
            Mar 5 at 15:23




            $begingroup$
            Thanks @Larry .
            $endgroup$
            – Felix Marin
            Mar 5 at 15:23


















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