### A New Year's Integral

###### 2012-12-07

A friend of mine showed me the following integral:

## Problem

#### Compute

$\int_0^1 dx\,\left[\left(1-x^{2011}\right)^{1⁄$2011}-x\right]^2.

## Solution

This is not the New Year’s integral of the title; clearly it would be a year out of date. That one will come a bit later.

My immediate thought when I saw this integral was that ${\left(1-{x}^{2011}\right)}^{1\mathrm{/}2011}$ looks like the usual thing whose limit one takes to get, or define, $e$. As $2011$ is pretty close to infinity, I numerically evaluated ${\int }_{0}^{1}dx\text{\hspace{0.17em}}{\left[{e}^{-x}-x\right]}^{2}$

and then numerically evaluated the real integral to see how they compared. Big mistake. First of all, it “looks like” it’s related to that limit involving $e$, but, more importantly, it isn’t! Second, the numerical evaluation of the actual integral gave too much away.

This integral numerically evaluates to $0.33333$, and leads you to wonder how $2011$, which isn’t even divisible by $3$, leads to an answer of $1\mathrm{/}3$. Replacing $2011$ by anything else and numerically evaluating again, we quickly find that the integral is completely independent of that parameter. So apparently we can just as well study $I={\int }_{0}^{1}dx\text{\hspace{0.17em}}{\left[{\left(1-{x}^{a}\right)}^{1\mathrm{/}a}-x\right]}^{2}\mathrm{.}$

Making the substitution $u=\left(1-{x}^{a}{\right)}^{1\mathrm{/}a},$

we find that $I=-{\int }_{0}^{1}du\text{\hspace{0.17em}}\frac{{u}^{a}\left(1-{u}^{a}{\right)}^{1\mathrm{/}a}}{u\left(1-{u}^{a}\right)}{\left[u-\left(1-{u}^{a}{\right)}^{1\mathrm{/}a}\right]}^{2}\mathrm{.}$

Now, wait a second, what if we tried a different substitution $y=\left(1-{x}^{a}{\right)}^{1\mathrm{/}a}-x$? We’d have $dy=dx\left(\frac{{x}^{a}\left(1-{x}^{a}{\right)}^{1\mathrm{/}a}}{x\left(1-{x}^{a}\right)}-1\right)\mathrm{.}$

Apparently if we take the two expressions for $I$ and add them, we have $2I={\int }_{0}^{1}dx\text{\hspace{0.17em}}\left(1-\frac{{x}^{a}\left(1-{x}^{a}{\right)}^{1\mathrm{/}a}}{x\left(1-{x}^{a}\right)}\right){\left[x-\left(1-{x}^{a}{\right)}^{1\mathrm{/}a}\right]}^{2}=-{\int }_{1}^{-1}dy\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}^{2}=\frac{2}{3}\mathrm{.}$

The answer is then that ${\int }_{0}^{1}dx\text{\hspace{0.17em}}{\left[{\left(1-{x}^{a}\right)}^{1\mathrm{/}a}-x\right]}^{2}=\frac{1}{3}$

regardless of $a$.

This is an interesting result for two reasons: (1) the integral has a parameter that looks pretty important (the $2011$) and turns out to not affect the answer in any way, and (2) the method appears to work by magic.

### Can we shed any light on this?

We can try to come up with a more general class of integrals that are susceptible to the same technique, but first let’s clarify exactly what the technique is. Suppose we have an integral $I={\int }_{a}^{b}dx\text{\hspace{0.17em}}f\left(x\right)$

whose integrand is symmetric under some coordinate transformation $x=g\left(u\right)$. By this, we mean that we pick an $f$ and a $g$ such that $f\left(x\right)=f\left(g\left(u\right)\right)=f\left(u\right)$ on some relevant domain. Then if we change variables, we find $I={\int }_{{g}^{-1}\left(a\right)}^{{g}^{-1}\left(b\right)}du\text{\hspace{0.17em}}\frac{dg}{du}f\left(u\right)\mathrm{.}$

Now, we’d like to combine the two expressions for $I$ which is easiest if they have the same bounds, so ${g}^{-1}\left(a\right)=a$ and ${g}^{-1}\left(b\right)=b$ or ${g}^{-1}\left(a\right)=b$ and ${g}^{-1}\left(b\right)=a$. For the moment, we’ll assume the former. If we impose this new condition, then we have $2I={\int }_{a}^{b}dx\text{\hspace{0.17em}}\left({g}^{\mathrm{\prime }}\left(x\right)+1\right)f\left(x\right)\mathrm{.}$

If we further require that $f\left(x\right)$ be only a function of $g\left(x\right)+x$, say $f\left(x\right)=h\left(g\left(x\right)+x\right)$, then a substitution $y=g\left(x\right)+x$ will solve the integral because $dy=dx\text{\hspace{0.17em}}\left({g}^{\mathrm{\prime }}\left(x\right)+1\right)\mathrm{.}$

But if we impose this additional restriction on $f$, then we get another piece of information, namely that $h\left(g\left(g\left(u\right)\right)+g\left(u\right)\right)=h\left(g\left(u\right)+u\right)\mathrm{.}$

One way of satisfying this is if $g\left(g\left(u\right)\right)=u$, so that $g\left(u\right)$ is an “involution”, a function that is its own inverse.

So, this meandering train of thought has brought us to one way of generalizing the original integral (of course, there are probably a few others). In summary,

1. Start with a function $g\left(u\right)$ which is its own inverse, $g\left(g\left(u\right)\right)=u$.
2. Find $a,b$ such that $g\left(a\right)=a$ and $g\left(b\right)=b$ or $g\left(a\right)=b$ and $g\left(b\right)=a$.
3. Pick $f\left(x\right)$ a function of $g\left(x\right)±x$, plus or minus as appropriate.

In the problem given at the beginning of this post, we have $g\left(u\right)=\left(1-{u}^{a}{\right)}^{1\mathrm{/}a}$ and $f\left(x\right)=\left[g\left(x\right)-x{\right]}^{2}$. We see that $g\left(g\left(u\right)\right)=u$, so it fits the mold, and $g\left(0\right)=1$ while $g\left(1\right)=0$.

### Happy New Year!

What are some examples of involutions? Well, a lot exist, but let’s pick a very simple-looking one: $g\left(x\right)=\frac{b}{x-a}+a\mathrm{.}$

What endpoints can we choose for an integral so that $g$ respects them? A little bit of tinkering shows that $g\left(1+a\right)=b+a$ while $g\left(b+a\right)=1+a$. Then we just need to pick $f$ a function of $g\left(x\right)-x$, the minus coming in because the endpoints get flipped around by $g$. If we let $c$ be a positive, even integer and take $f\left(x\right)={\left[g\left(x\right)-x\right]}^{c},$

then we will have ${\int }_{1+a}^{a+b}dx\text{\hspace{0.17em}}{\left(\frac{b}{x-a}+a-x\right)}^{c}=\frac{1}{2\left(c+1\right)}\left(\left(b-1{\right)}^{c+1}-\left(1-b{\right)}^{c+1}\right)$

by the above method.

Finally, letting $a=-1$, $b=2$ and $c=2012$, we have the deceptively simple-looking ${\int }_{0}^{1}dx\text{\hspace{0.17em}}{\left(\frac{2}{x+1}-1-x\right)}^{2012}=\frac{1}{2013}\mathrm{.}$

Mathematica 8 appears unable to handle this integral, even giving the wrong answer numerically under default precision settings!