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

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

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

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?

• Perhaps the simpler integral (with $r=1$) can be used in an integration-by-parts approach: $\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}-\frac{1}{2}$ – David G. Stork Jan 11 at 23:45
• 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 $0$. – lulu Jan 12 at 0:03
• And, sticking with the header notation and letting $n=2$, we see that $\int_0^1 ((1-x^2)^{1/2}-x)^3\,dx = \frac {3\pi}8 -1 \approx .1781 \neq \frac 14$ . Or am I misreading something? – lulu Jan 12 at 0:08
• I see, thank you for re-asking the question! It is a nice integral for sure. – Zacky Jan 12 at 0:26
• @clathratus: Oops, $u$ should be $y$. I will edit it. Thanks for pointing that out. – Larry Jan 12 at 0:51

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

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

• Beautiful! This looks like Glasser's Master theorem little brother :D – Zacky Jan 12 at 1:02
• Was it necessary to start with absolute value in the argument of function? – user Jan 12 at 1:04
• @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 Jan 12 at 1:06
• @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.) – Sangchul Lee Jan 12 at 1:12
• @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... – Sangchul Lee Mar 17 at 7:37

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

• Try to find an error in your derivation as $I(r,n)=\frac{1}{n+1}$ for even $n$ and any $r$. – user Jan 12 at 1:45
• @user Ah yes I forgot $n$ was even – clathratus Jan 12 at 1:57
• Nice approach. Seems almost obvious after reading it, but I was stuck when I first saw it (+1) – user150203 Jan 12 at 5:31

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


• Nice solution, (+1). – Larry Mar 5 at 13:19
• Thanks @Larry . – Felix Marin Mar 5 at 15:23