Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A continuous functions you can't integrate?

  1. Dec 8, 2006 #1

    Alkatran

    User Avatar
    Science Advisor
    Homework Helper

    After reading http://en.wikipedia.org/wiki/Weierstrass_function it occurred to me that I could do the same thing to an integral:

    [tex]
    \int \sum_{i=0}^\infty \frac{sin(\frac{x}{3^i})}{2^i} dx
    [/tex]

    [tex]
    = \sum_{i=0}^\infty \int \frac{sin(\frac{x}{3^i})}{2^i} dx
    [/tex]

    [tex]
    = \sum_{i=0}^\infty -cos(\frac{x}{3^i})(\frac{3}{2})^i
    [/tex]

    which obviously diverges.

    If we turn this into a definite integral and take the limit as n goes to infinity, it becomes a conditionally convergent sum. So we have the odd property that the indefinite integral doesn't exist, but the definite one does.

    [tex]
    \lim_{n \rightarrow \infty} \int_a^b \sum_{i=0}^n \frac{sin(\frac{x}{3^i})}{2^i} dx
    [/tex]

    [tex]
    = \lim_{n \rightarrow \infty} \sum_{i=0}^n (cos(\frac{a}{3^i}) - cos(\frac{b}{3^i}))(\frac{3}{2})^i dx
    [/tex]

    [tex]
    = \lim_{n \rightarrow \infty} \sum_{i=0}^n -2sin(\frac{a+b}{2*3^i})sin(\frac{a - b}{2*3^i})(\frac{3}{2})^i dx
    [/tex]
    which should converge because sin(x) becomes linear as x approaches 0.

    What exactly is the explanation for this? I thought all continuous functions could be integrated (they definitely told us that!).
     
    Last edited: Dec 8, 2006
  2. jcsd
  3. Dec 8, 2006 #2

    StatusX

    User Avatar
    Homework Helper

    Indefinite integrals are just definite integrals where one of the bounds, say the upper bound, is a variable. We can usually just ignore the lower bound completely, since starting at a different point just amounts to adding an overall constant, which we don't care about. This would even be fine when you have a finite sum of integrals, ie, ignoring the bounds would be the same as starting each integral at possibly different lower bounds, which isn't strictly correct, but just adds to the final answer a finite sum of constants. But in the case of an infinite sum of integrals, you are shifting the answer by a constant for each term, and the sum of these constant turns out in this case to diverge. Note you can just replace b by x in the second part, change it by a constant if you want, and you have the indefinite integral.
     
    Last edited: Dec 8, 2006
  4. Dec 8, 2006 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What justification do you have for supposing that this swapping of summation and integral is at all valid? Clearly, in general it is not a valid operation.
     
  5. Dec 8, 2006 #4

    Alkatran

    User Avatar
    Science Advisor
    Homework Helper

    I didn't know that. What kind of situations is it not valid in, besides this one? What if took the limit as n goes to infinity, could I switch it then?
     
  6. Dec 8, 2006 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Erm, that infinite sum is *by definition* the limit as 'n tends to infinity'.

    It is the standard difference between uniform and conditional convergence.
     
  7. Dec 8, 2006 #6

    StatusX

    User Avatar
    Homework Helper

    The series converges absolutely for all x, so I think exchaning the integral and sum as in the second half of post 1 is valid.
     
  8. Dec 9, 2006 #7

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    I can't say I checked the post very carefully. There are two possible errors. One, is that swapping summation and integration needs justification. The second is: when you did that indefinite integral, each integral comes with a 'plus constant'. Let's call integral F(x). What you appear to be saying is that if we work out F(b)-F(a) by taking limits, then for any a,b this gives a finite answer. But if I just work out F(x) for any x by taking limits I get infinity. That is not very surprising: it is like splitting the sum in the last but one line of maths in your post into two sums. In general you cannot swap the order of summation on things.
     
    Last edited: Dec 9, 2006
  9. Dec 9, 2006 #8

    Gib Z

    User Avatar
    Homework Helper

    I'm not sure, but i've heard that some functions do not have indefinite integrals, and can only be approximated. Or maybe the authors meant no integral in elementary functions..If true Im thinking fractals...Think of a Koch Snowflake, the definite integral converges, but can not be differentiated, and therefore not integrated, anywhere. Am i correct?
     
  10. Dec 9, 2006 #9

    Alkatran

    User Avatar
    Science Advisor
    Homework Helper

    No, you can integrate some functions that you can't differentiate. See the function in the wikipedia article in my first post: it can be integrated, but not differentiated.
     
  11. Dec 13, 2006 #10

    Alkatran

    User Avatar
    Science Advisor
    Homework Helper

    That's a different situation. For example, the integral of e to the x squared can't be expressed as an elementary function, even though e^(x^2) is an elementary function. You can, however, express it as an infinite sum (I think), but that sum converges absolutely. In this case the sum converges conditionally.

    See this wikipedia article: http://en.wikipedia.org/wiki/Elementary_function
     
  12. Dec 13, 2006 #11

    Mute

    User Avatar
    Homework Helper

    Does it converge uniformly, though?

    You can say it converges along the real axis because

    [tex] \sum_{i=0}^\infty \frac{sin(\frac{x}{3^i})}{2^i}[/tex]

    is clearly bounded by

    [tex] \sum_{i=0}^\infty \frac{1}{2^i}[/tex],

    which converges to 2.

    However, if you consider

    [tex] \sum_{i=0}^\infty \frac{sin(\frac{z}{3^i})}{2^i}[/tex]

    with z complex, then you can't bound the series because sin(z) grows arbitrarily large along the imaginary axis.

    So, maybe I'm missing something , but it looks to me like this series doesn't converge uniformly on the entire complex plane, which would prohibit the swapping of the integral and summation. (Of course, I could be wrong here, given that Fourier series like to converge and I could probably make the same argument there, but we spent a whole semester calling those uniformly convergent in physics, so I may very well be missing something here.)
     
    Last edited: Dec 13, 2006
  13. Dec 13, 2006 #12

    StatusX

    User Avatar
    Homework Helper

    By the dominated convergence theorem, you can swap the integral and sum (or more specifically, the integral and the limit of the finite sums) if there is an integrable nonnegative function that bounds each of the partial sums. Since:

    [tex]\sum_{i=0}^N \left|\frac{sin(\frac{x}{3^i})}{2^i}\right| \leq \sum_{i=0}^N \frac{1}{2^i}< 2[/tex]

    2 is such a function.
     
    Last edited: Dec 13, 2006
  14. Dec 14, 2006 #13

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    I think the assumption, Mute, was that we were only considering this as a function of a real variable.
     
  15. Dec 14, 2006 #14

    Mute

    User Avatar
    Homework Helper

    Yes, I realized that, but I was thinking along the lines of how the series for 1/(1+x^2) didn't converge despite having no singularities on the Real axis, but due to the fact it had one on the imaginary axis. I reasoned that perhaps this series might not be uniformly convergent on the complex plane which might limit operations we can do on it with just real calculus. I wasn't sure, though, as I sort of stated at the end, but I figured I might as say something, because at least if I was wrong, someone would correct me and I might learn something. =P
     
  16. Dec 27, 2006 #15
    think "Alkatran" has avoided an important feature..we can't alwys take the sum of an integral equal to the integral of the sums..it's strange but true, the same happens with Borel resummation...if series is convergent you can take term-by-term Laplace transform, else (if divergent) you can't do it...
     
  17. Dec 27, 2006 #16

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    As had been pointed out before, several times, and explained, and the real problem identified. Still, nice of you to try to bring it back round to your pet topic, Jose. Can we expect a post berating mathematicians for being to snobbish to let this function be integrated?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?