A friend of mine showed me the following integral:
This is not the New Year’s integral of the title; clearly it would be a year out of date.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 looks like the usual thing whose limit one takes to get, or define, . As is pretty close to infinity, I numerically evaluated
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 , but, more importantly, it isn’t! Second, the numerical evaluation of the actual integral gave too much away.
This integral numerically evaluates to , and leads you to wonder how , which isn’t even divisible by , leads to an answer of . Replacing 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
Making the substitution
we find that
Now, wait a second, what if we tried a different substitution ? We’d have
Apparently if we take the two expressions for and add them, we have
The answer is then that
regardless of .
This is an interesting result for two reasons: (1) the integral has a parameter that looks pretty important (the ) 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
whose integrand is symmetric under some coordinate transformation . By this, we mean that we pick an and a such that on some relevant domain. Then if we change variables, we find
Now, we’d like to combine the two expressions for which is easiest if they have the same bounds, so and or and . For the moment, we’ll assume the former. If we impose this new condition, then we have
If we further require that be only a function of , say , then a substitution will solve the integral because
But if we impose this additional restriction on , then we get another piece of information, namely that
One way of satisfying this is if , so that 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,
In the problem given at the beginning of this post, we have and . We see that , so it fits the mold, and while .
What are some examples of involutions? Well, a lot exist, but let’s pick a very simple-looking one:
What endpoints can we choose for an integral so that respects them? A little bit of tinkering shows that while . Then we just need to pick a function of , the minus coming in because the endpoints get flipped around by . If we let be a positive, even integer and take
then we will have
by the above method.
Finally, letting , and , we have the deceptively simple-looking
Mathematica 8 appears unable to handle this integral, even giving the wrong answer numerically under default precision settings!