A friend of mine showed me the following integral:

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

This is not the New Year’s integral of the title; clearly it would be a year out of date.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 ${(1-{x}^{2011})}^{1\mathrm{/}2011}$ looks like the usual thing whose limit one takes to get, or define, $e$. As $2011$ is pretty close to infinity, I numerically evaluated ${\int}_{0}^{1}dx\text{\hspace{0.17em}}{[{e}^{-x}-x]}^{2}$

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 $I={\int}_{0}^{1}dx\text{\hspace{0.17em}}{[{(1-{x}^{a})}^{1\mathrm{/}a}-x]}^{2}\mathrm{.}$

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

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

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

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

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

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.

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 $f\left(x\right)=f\left(g\right(u\left)\right)=f\left(u\right)$ 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 $2I={\int}_{a}^{b}dx\text{\hspace{0.17em}}\left({g}^{\mathrm{\prime}}\right(x)+1)f\left(x\right)\mathrm{.}$

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

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

One way of satisfying this is if $g\left(g\right(u\left)\right)=u$, 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,

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

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

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 $g(1+a)=b+a$ while $g(b+a)=1+a$. 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 $f\left(x\right)={\left[g\right(x)-x]}^{c},$

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

by the above method.

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

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