On 2008 Feb 25, James Propp <jpropp@cs.uml.edu> wrote:
Are there any theories of integration out there that would make the function
{-(2 Cos[1/x^2])/x + 2 x Sin[1/x^2] for x not equal to 0, f(x) = { { 0 for x = 0
integrable on the interval [-1,1]?
Yes. (Sorry for my late response. Also sorry if someone has already mentioned this and I overlooked it.) It's gauge integrable. See, for example, <http://en.wikipedia.org/wiki/Henstock-Kurzweil_integral>. So did Jim's post reveal a "dirty secret of calculus"? I would say so. But the secret it reveals, IMO, is simply that the Riemann integral should be replaced by the gauge integral in introductory calculus courses. David W. Cantrell
Note that the function f(x), although discontinuous at x=0, is equal to the derivative of the function
{x^2 Sin[1/x^2] for x not equal to 0, F(x) = { { 0 for x = 0
for ALL x (including x = 0). So, if we were to blindly apply the Fundamental Theorem of the Calculus without attending to the satisfaction of its hypotheses, we would conclude that the integral of f from -1 to 1 equals F(1) - F(-1). Proceeding less blindly, it seems safe to say that any theory of integration that assigned a value to the integral would have to assign it the value F(1) - F(-1). But the unbounded behavior of f(x) in the vicinity of x=0 makes things difficult. Indeed, the function f(x) is neither Riemann integrable nor Lebesgue integrable on [-1,1]. See http://jamespropp.org/142/FunctionsBehavingWorse.nb for a sketch of the graph.
(Can nonstandard analysis help here? For some reason I have a feeling that the Stone-Cech compactification might be of use.)
Another question arising from this example is: Can things be even worse? Say a function is locally unbounded at a if it is unbounded on every neighborhood of a. If a function F is differentiable everywhere, what can be said about the points at which the derivative is locally unbounded? Can it be dense? Can it be all of R?