A New Year's Integral
A friend of mine showed me the following integral: ## Problem #### Compute \begin{equation*} \int_0^1 dx\,\left[\left(1-x^{2011}\right)^{1/2011}-x\right]^2. \end{equation*} ## Solution This is not the New Year's integral of the title; clearly it would be a year out of date.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 $\left(1-x^{2011}\right)^{1/2011}$ looks like the usual thing whose limit one takes to get, or define, $e$. As $2011$ is pretty close to infinity, I numerically evaluated $$\int_0^1 dx\,\left[e^{-x}-x\right]^2$$ 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 $1/3$. 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 $$I=\int_0^1 dx\,\left[\left(1-x^a\right)^{1/a}-x\right]^2.$$ Making the substitution $$u=(1-x^a)^{1/a},$$ we find that $$I=-\int_0^1 du\,\frac{u^a(1-u^a)^{1/a}}{u(1-u^a)}\left[u-(1-u^a)^{1/a}\right]^2.$$ Now, wait a second, what if we tried a different substitution $y=(1-x^a)^{1/a}-x$? We'd have $$dy=dx\left(\frac{x^a(1-x^a)^{1/a}}{x(1-x^a)}-1\right).$$ Apparently if we take the two expressions for $I$ and add them, we have $$2I=\int_0^1 dx\,\left(1-\frac{x^a(1-x^a)^{1/a}}{x(1-x^a)}\right)\left[x-(1-x^a)^{1/a}\right]^2=-\int_1^{-1} dy\,\,y^2=\frac{2}{3}.$$ The answer is then that $$\int_0^1 dx\,\left[\left(1-x^a\right)^{1/a}-x\right]^2=\frac{1}{3}$$ 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 $$I=\int_a^b dx\,f(x)$$ 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 $$I=\int_{g^{-1}(a)}^{g^{-1}(b)} du\,\frac{dg}{du} f(u).$$ 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 $$2I=\int_a^b dx\,\left(g'(x)+1\right)f(x).$$ 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 $$dy=dx\,\left(g'(x)+1\right).$$ But if we impose this additional restriction on $f$, then we get another piece of information, namely that $$h\Big(g(g(u))+g(u)\Big)=h(g(u)+u).$$ One way of satisfying this is if $g(g(u))=u$, so that $g(u)$ is an "[involution](http://en.wikipedia.org/wiki/Involution_(mathematics))", 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 $g(u)$ which is its own inverse, $g(g(u))=u$. 2. Find $a,b$ such that $g(a)=a$ and $g(b)=b$ or $g(a)=b$ and $g(b)=a$. 3. 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 $g(u)=(1-u^a)^{1/a}$ and $f(x)=[g(x)-x]^2$. We see that $g(g(u))=u$, so it fits the mold, and $g(0)=1$ while $g(1)=0$. ### Happy New Year! What are some examples of involutions? Well, [a lot exist](http://math.stackexchange.com/questions/46635/examples-of-involutions-on-mathbbr), but let's pick a very simple-looking one: $$g(x)=\frac{b}{x-a}+a.$$ 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 $$f(x)=\left[g(x)-x\right]^c,$$ then we will have $$\int_{1+a}^{a+b} dx\,\left(\frac{b}{x-a}+a-x\right)^c=\frac{1}{2(c+1)}\left((b-1)^{c+1}-(1-b)^{c+1}\right)$$ by the above method. Finally, letting $a=-1$, $b=2$ and $c=2012$, we have the deceptively simple-looking $$\int_0^1 dx\,\left(\frac{2}{x+1}-1-x\right)^{2012}=\frac{1}{2013}.$$ Mathematica 8 appears unable to handle this integral, even giving the wrong answer numerically under default precision settings!