A New Year's Integral

2012-12-07 math integral

A friend of mine showed me the following integral:



01dx[(1x2011)1/2011x]2. \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This is not the New Year’s integral of the title; clearly it would be a year out of date.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 (1x2011)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 01dx[exx]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=01dx[(1xa)1/ax]2.

Making the substitution u=(1xa)1/a,

we find that I=01duua(1ua)1/au(1ua)[u(1ua)1/a]2.

Now, wait a second, what if we tried a different substitution y=(1xa)1/ax? We’d have dy=dx(xa(1xa)1/ax(1xa)1).

Apparently if we take the two expressions for I and add them, we have 2I=01dx(1xa(1xa)1/ax(1xa))[x(1xa)1/a]2=11dyy2=23.

The answer is then that 01dx[(1xa)1/ax]2=13

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=abdxf(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=g1(a)g1(b)dudgduf(u).

Now, we’d like to combine the two expressions for I which is easiest if they have the same bounds, so g1(a)=a and g1(b)=b or g1(a)=b and g1(b)=a. For the moment, we’ll assume the former. If we impose this new condition, then we have 2I=abdx(g(x)+1)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(g(x)+1).

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).

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)±x, plus or minus as appropriate.

In the problem given at the beginning of this post, we have g(u)=(1ua)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, but let’s pick a very simple-looking one: g(x)=bxa+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)=[g(x)x]c,

then we will have 1+aa+bdx(bxa+ax)c=12(c+1)((b1)c+1(1b)c+1)

by the above method.

Finally, letting a=1, b=2 and c=2012, we have the deceptively simple-looking 01dx(2x+11x)2012=12013.

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

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