A New Year's Integral

2012-12-07 math integral

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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This is not the New Year’s integral of the title; clearly it would be a year out of date.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 ${(1-x^{2011})}^{1\mathrm{/}2011}$ looks like the usual thing whose limit one takes to get, or define, $e$. As $2011$ is pretty close to infinity, I numerically evaluated $${\int }_0^{1}dx{[e^{-x}-x]}^2$$

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 $$I={\int }_0^{1}dx{[{(1-x^a)}^{1\mathrm{/}a}-x]}^2\mathrm{.}$$

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

we find that $$I=-{\int }_0^{1}du\frac{u^a(1-u^a{)}^{1\mathrm{/}a}}{u(1-u^a)}{[u-(1-u^a{)}^{1\mathrm{/}a}]}^2\mathrm{.}$$

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 $$2I={\int }_0^{1}dx(1-\frac{x^a(1-x^a{)}^{1\mathrm{/}a}}{x(1-x^a)}){[x-(1-x^a{)}^{1\mathrm{/}a}]}^2=-{\int }_1^{-1}dy{y}^2=\frac{2}{3}\mathrm{.}$$

The answer is then that $${\int }_0^{1}dx{[{(1-x^a)}^{1\mathrm{/}a}-x]}^2=\frac{1}{3}$$

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.

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 $$I={\int }_a^{b}dxf(x)$$

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 $$I={\int }_{g^{-1}(a)}^{g^{-1}(b)}du\frac{dg}{du}f(u)\mathrm{.}$$

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 $$2I={\int }_a^{b}dx(g^{\mathrm{\prime }}(x)+1)f(x)\mathrm{.}$$

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 $$dy=dx(g^{\mathrm{\prime }}(x)+1)\mathrm{.}$$

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,

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 $g(u)=(1-u^a{)}^{1\mathrm{/}a}$ and $f(x)=[g(x)-x{]}^2$. We see that $g(g(u))=u$, so it fits the mold, and $g(0)=1$ while $g(1)=0$.

Happy New Year!

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 $${\int }_{1+a}^{a+b}dx{(\frac{b}{x-a}+a-x)}^c=\frac{1}{2(c+1)}((b-1{)}^{c+1}-(1-b{)}^{c+1})$$

by the above method.

Finally, letting $a=-1$, $b=2$ and $c=2012$, we have the deceptively simple-looking $${\int }_0^{1}dx{(\frac{2}{x+1}-1-x)}^{2012}=\frac{1}{2013}\mathrm{.}$$

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