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View Full Version : Why are some functions that aren't integratable?


Netbrian
12-12-2002, 07:18 PM
Currently in math class, we've been calculating the distance of curves in 3D space (how exciting!) One such function was given, and after we started calculating, it took my rather powerful TI89 several minutes to actually DO. We then glossed over functions that are impossible to integrate (wasn't sure if this one was one of those and the calculator just calculated an insane amount of tiny increments to get an approximate solution), but didn't go further than that. What I'm wonder is, what makes something completely unintegratable, why is it unintegratable, and is there any chance of such functions being integrated with some new mathematics in the future?

Thanks!

MC Master of Ceremonies
12-12-2002, 07:30 PM
I feel I should know the answer to this one, because at some vague point I was doing a physics degree (though I changed to a completely unrelated subject).

I'd just ask this: what type of functions were you intergrating?

Jabba
12-12-2002, 07:45 PM
There are 2 ways a function may be described as "not integrable", depending on the context.
Firstly, there is no function f(x) such that f '(x) = e-x2, so e-x2 has no indefinite integral. However this is not what is meant by integrable in a maths textbook and the above function would be classed as integrable.

Non-integrable in an analysis textbook means that the function has no definite integral. This could be
(i) because the function is too big. For example, x is not integrable on R.
(ii) because the function is nasty. This is a more difficult idea and the amount of nastiness required depends on which integral you are using. For example, define the function f by: f(x) is 1 whenever x is rational and f(x) is 0 whenever x is irrational. Then f is not Riemann integrable but it is Lebesgue integrable ( over R) with integral 0.

Lebesgue was able to prove:
Let f be defined and bounded on [a,b] and let D be the set of discontinuities of f in [a,b]. Then f is Riemann integrable on [a,b] if and only if D has measure zero.

Jabba
12-12-2002, 07:51 PM
Sorry, in the above I meant " there is no function f with a simple closed formula such that f ' (x) = ..."

David Simmons
12-12-2002, 10:45 PM
My WAG. In order for a function to have an integral, the sum of the increments f(x)dx (dx represents delta x) must have a limit as dx approaches zero and the number of increments increases without limit.

Pythagras
12-12-2002, 10:49 PM
Is there an analytical solution for the integral of x^x from 0 to x dx?

ultrafilter
12-12-2002, 10:49 PM
In addition, some functions are bad cases for the TI-89's approximation algorithm, simply because they're hard to compute, and you do have to do a lot of values to get a reasonable approximation.

There are functions which are not Lebesgue integrable, but Lebesgue integration is almost a century old, and it's certainly not the newest method. To the best of my knowledge, the most general integral out there is due to McShane, but it's pretty advanced--you actually have to understand quite a bit of probability theory to understand it.

ultrafilter
12-12-2002, 10:52 PM
Strike that--the McShane integral arose out of probability theory, but I don't think it's inherently probabilistic.

By the way, I'm pretty sure that the only reason a measurable function can fail to be integrable is if its integral diverges.

ultrafilter
12-12-2002, 10:55 PM
Originally posted by David Simmons
My WAG. In order for a function to have an integral, the sum of the increments f(x)dx (dx represents delta x) must have a limit as dx approaches zero and the number of increments increases without limit.

Pretty much. The limit also has to be the same no matter which point you pick from each interval.