By Larry


2019-01-11 23:26:40 8 Comments

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 $\left\{n = 2k,\ r > 0\ \mid\ k \in \mathbb{N}\right\}$

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

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?

4 comments

@Felix Marin 2019-03-05 01:10:34

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

@Larry 2019-03-05 13:19:36

Nice solution, (+1).

@Felix Marin 2019-03-05 15:23:44

Thanks @Larry .

@Sangchul Lee 2019-01-12 00:53:59

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 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,

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

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

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

@Zacky 2019-01-12 01:02:48

Beautiful! This looks like Glasser's Master theorem little brother :D

@user 2019-01-12 01:04:12

Was it necessary to start with absolute value in the argument of function?

@Sangchul Lee 2019-01-12 01:06:31

@user, It is kind of necessary, in the sense that $$ \int_{0}^{1} g\left(x-(1-x^r)^{1/r}\right) \, \mathrm{d}x = \int_{0}^{1} g(u) \, \mathrm{d}u $$ may fail if $g$ is not an even function on $[-1, 1]$.

@Sangchul Lee 2019-01-12 01:12:11

@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.)

@Larry 2019-01-27 23:17:02

@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.

@Sangchul Lee 2019-01-27 23:49:59

@Larry, if $n$ is a positive even integer, then $a^n = (-a)^n = |a|^n$.

@ersh 2019-03-17 05:04:52

@Sangchul Lee, Amazing!!! I feel ashamed of my maths.

@Sangchul Lee 2019-03-17 07:37:33

@ersh, Thank you, and don't feel ashamed :) I was just lucky enough to find these approaches. It is like seeing facebook postings, which are basically a collage of someone else's finest moments...

@ablmf 2019-01-12 11:03:22

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 $\frac 1 {2n+1}$.

@clathratus 2019-01-12 00:59:24

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

@user 2019-01-12 01:45:11

Try to find an error in your derivation as $I(r,n)=\frac{1}{n+1}$ for even $n$ and any $r$.

@clathratus 2019-01-12 01:57:23

@user Ah yes I forgot $n$ was even

@DavidG 2019-01-12 05:31:40

Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1)

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