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What you did in the limit in your second expression does not make any sense. You can't obtain the indefinite integral (the expression with cos()) and then take
by mturmon 13y ago
What you did in the limit in your second expression does not make any sense. You can't obtain the indefinite integral (the expression with cos()) and then take the limit as n -> infinity. This is not meaningful.
You can only do that with a definite integral (under conditions).
Using your reasoning, even a function like exp(x) would not integrate, because the terms in the integral of its series expansion would all go to zero as n -> infinity.
Another way to think about this is that each term in the indefinite integral you wrote has an arbitrary constant of integration, which can depend on n, that you did not include. This constant precludes you from actually finding the limit.
- Strilanc 13y agoI mentioned that I was playing fast and loose when swapping them. But nevermind that exact process, let's be a bit more direct. 1. Can you give me any function whose derivative is the function I specified? (i.e. I give you permission to force all the integration constants to zero) 2. Can you compute the area under the curve from `x1` to `x2` of the function I specified? 3. Do you agree that you can do (2) but not (1)?
- deleted 13y ago[deleted]
- mturmon 13y agoI see what you're getting at now. Let's check the conditions for integrability. Your function is bounded (everywhere, but let's restrict to [0,1]). But I'm not sure it is continuous due to the limit process. If you can't show it's continuous almost everywhere (as noted by @clintonc) then it is not Riemann-integrable, therefore its integral will not exist. That would be the answer to (1). If you go through some manipulations (i.e., (2)) and get an expression that seems correct, but the underlying function is not actually Riemann-integrable (as in (1)), then I'd suggest your manipulations have fooled you. So, do you know the answers to these questions?
- Strilanc 13y agoThe function is continuous. The cumulative effect of the sum after the nth item are bounded by a [-2^-n, 2^-n] offset, and varies very slowly as x is changed (the argument to cos is divided by increasingly huge factors). I haven't done an explicit epsilon-delta proof, but I'm confident I could make one work by using those two facts. So I don't expect to be able to show lack of Riemann-integrable-ness via that route. ... I just don't think it has an antiderivative. ... wait, this makes no sense. I've made a mistake because if the function is continuous then it must have an antiderivative by the first fundamental theorem of calculus. I will have to actually do the epsilon-delta proof and find out where I made the tricky mistake. Probably something to do with that infinity...
- SamReidHughes 13y agoThe series (of continuous functions) converges uniformly, which means that it's Reimann integrable. And you can compute its integral as a sum of the integrals inside the sum (which you showed upthread). Also, as hinted upthread, just change f(x) = - sum[n = 0 to infinity](10^n / pi cos((1/20)^n pi x)) to f(x) = - sum[n = 0 to infinity](10^n / pi (cos((1/20)^n pi x) - 1)) That gives you an actual formula for the antiderivative that converges at every point.
- mturmon 13y agoBingo. Very clear.