A friend of mine showed me the following integral:

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

This is not the New Year’s integral of the title; clearly it would be a year out of date.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
$$I={\int}_{a}^{b}dx\text{\hspace{0.17em}}f(x)$$

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

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

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

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

One way of satisfying this is if $g(g(u))=u$, so that $g(u)$ 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 $g(u)$ which is its own inverse, $g(g(u))=u$.
- Find $a,b$ such that $g(a)=a$ and $g(b)=b$ or $g(a)=b$ and $g(b)=a$.
- Pick $f(x)$ a function of $g(x)\pm x$, plus or minus as appropriate.

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

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

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 $g(x)-x$, the minus coming in because the endpoints get flipped around by $g$. If we let $c$ be a positive, even integer and take $$f(x)={[g(x)-x]}^{c},$$

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

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!