4 ms·

Not sure what you mean by this, since your comment has no context -- there are no limits or sums in the link. However, even the humble Riemann integral would b

Not sure what you mean by this, since your comment has no context -- there are no limits or sums in the link. However, even the humble Riemann integral would be able to find the area under the graph of a bounded function on a compact interval which is continuous on a set of full Lebesgue measure. Any other function only has a "well-defined area" for some seriously stunted notion of area; you can integrate more functions with a different integral, but what one calls a "well-defined area" pretty much depends on what kind of integral you're using at that point.

Strilanc 12y ago
Consider this function: f(x) = sum[n = 0 to infinity]((1/2)^n sin((1/20)^n pi x)) It converges because each successive term is bounded by [-1/2, +1/2]. But if you integrate it you get: f(x) = sum[n = 0 to infinity](10^n / pi cos((1/20)^n pi x)) (Well... you get that if you play fast and loose about swapping the order of the integral and the sum) Which diverges because the input to the cosine function limits to 0 as n -> infinity, so the cos limits to 1, and we get 10+100+1000+10000... But if you integrate from two points and delay doing the limit until after then you get F(x1, x2) = sum[n = 0 to infinity](10^n / pi (cos((1/20)^n pi x2) - cos((1/20)^n pi x1))) = sum[n = 0 to infinity](10^n / pi (2*sin(pi/2 (1/20)^n (x1-x2))*sin(pi/2 (1/20)^n (x1+x2))) Which, for large n, acts like: ~= sum[n = 0 to infinity](10^n / pi (2*(pi/2 (1/20)^n (x1-x2))*(pi/2 (1/20)^n (x1+x2))) = sum[n = 0 to infinity]((1/40)^n pi/2 *(x1-x2)(x1+x2)) Which converges. Apologies for any math mistakes. This was all off the cuff. I wouldn't be surprised if some of the more general integration or summation strategies can also handle this case.
- mturmon 12y 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 12y ago
    I 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 12y ago
      [deleted]
    - mturmon 12y ago
      I 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 12y ago
        The 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...

Strilanc 12y ago

Consider this function: f(x) = sum[n = 0 to infinity]((1/2)^n sin((1/20)^n pi x)) It converges because each successive term is bounded by [-1/2, +1/2]. But if you integrate it you get: f(x) = sum[n = 0 to infinity](10^n / pi cos((1/20)^n pi x)) (Well... you get that if you play fast and loose about swapping the order of the integral and the sum) Which diverges because the input to the cosine function limits to 0 as n -> infinity, so the cos limits to 1, and we get 10+100+1000+10000... But if you integrate from two points and delay doing the limit until after then you get F(x1, x2) = sum[n = 0 to infinity](10^n / pi (cos((1/20)^n pi x2) - cos((1/20)^n pi x1))) = sum[n = 0 to infinity](10^n / pi (2*sin(pi/2 (1/20)^n (x1-x2))*sin(pi/2 (1/20)^n (x1+x2))) Which, for large n, acts like: ~= sum[n = 0 to infinity](10^n / pi (2*(pi/2 (1/20)^n (x1-x2))*(pi/2 (1/20)^n (x1+x2))) = sum[n = 0 to infinity]((1/40)^n pi/2 *(x1-x2)(x1+x2)) Which converges. Apologies for any math mistakes. This was all off the cuff. I wouldn't be surprised if some of the more general integration or summation strategies can also handle this case.

mturmon 12y 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 12y ago
  I 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 12y ago
    [deleted]
  - mturmon 12y ago
    I 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 12y ago
      The 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...