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I General questions about integrals (definitions)

  1. Apr 12, 2016 #1
    Hi everybody! I'm currently studying integrals, and I would like to clarify a few definitions, especially about the criterions of convergence/divergence of an integral. Basically if that's okay for you guys I'm gonna list and number a few statements and I'd like to know if they are true or not.

    1. If a definite integralab f(x) dx has both its domain of integration and integrand bounded, but its set of locations where it is not continuous is an uncountable set, then f(x) is not integrable.
    (or do we have to run other tests to find out if the function is integrable or not?)

    2. If a definite integralab f(x) dx has both its domain of integration and integrand unbounded, then it is called an improper integral. While the term improper integral normally designates the limit of an unbounded integral, this "ambiguity is resolved as both the proper and improper integral will coincide in value" (Wikipedia, improper integral).

    3. The improper integral of a function f(x) exists, if there exists a function g(x) so that ∫ab g(x) dx converges and |f(x)| ≤ g(x) ∀ x ∈ [a,b). If one of those criterions is not met, then we cannot conclude anything about the existence of the improper integral of f(x) and we must run other tests. (Majorant criterion)

    4. The improper integral of a function f(x) doesn't exist, if there exists a function g(x) so that ∫ab g(x) dx diverges and f(x) ≥ g(x) ∀ x ∈ [a,b). If one of those criterions is not met, then we cannot conclude anything about the non-existence of the improper integral of f(x) and we must run other tests. (Minorant criterion)


    I stop here, it would already mean a lot to me if those simple assumptions would become facts. :)

    Thank you very much in advance, I appreciate your help.


    Julien.
     
  2. jcsd
  3. Apr 12, 2016 #2

    Samy_A

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    For 1), consider the function ##f: \mathbb R \to \mathbb R## defined by:
    ##\left\{\begin{array}{l}
    0 \ on \ \mathbb Q\\
    1 \ on \ \mathbb R \setminus \mathbb Q
    \end{array}
    \right.##

    For 4), consider ##g(x)=-\frac{1}{x}## and ##f(x)=0## on ##]0,1[##
     
  4. Apr 12, 2016 #3
    @Samy_A Hi and thank you very much for your answers.

    It seems to me that this function is integrable though its set of non continuous locations is uncountable. Is that right? Then I would conclude that my statement was false and that this criterion only lets us know if a function is integrable but not if it is not integrable.

    I see, f(x) converges though it is bigger as g(x).. Would that problem be solved if I had stated |f(x)| ≥ g(x)?

    Oh, and also are 2. and 3. correct statements then?

    Thank you so much Samy.

    Julien.
     
  5. Apr 12, 2016 #4
    Another question related to 1. would then be: if f(x) is bounded on both its domain of integrability and integrand but its set of non continuous locations is uncountable, what criterion(s) am I left with to check if f(x) is integrable or not?


    Julien.
     
  6. Apr 12, 2016 #5

    Samy_A

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    Statement 1) is indeed false. I don't exactly understand what criterion for being integrable or not it could yield.
    No, see the same counterexample.
    3) is correct (provided f is a measurable function). I'm not sure about what 2) is supposed to mean.
    That's a difficult question to answer. It depends on how you studied integration: Riemann integral, Lebesgue integral?
     
  7. Apr 12, 2016 #6
    @Samy_A Thank you again for your answers, I appreciate your help.

    Mmm that comes from the script of my maths teacher. It's in German but I try to translate it:

    Bounded functions are integrable, if they are "almost everywhere" continuous: a bounded function f: [a,b] → ℂ is integrable, if the set of its discontinuities is countable.

    Oh yes indeed. Maybe f(x) ≥ |g(x)| then? :-p Or is it that the minorant criterion does not exist at all for integration?

    About 2: I was under the impression that what we called an improper integral is actually the limit of a definite integral as one of the boundaries tends towards ±∞ or as the integrand tends towards ±∞. Actually I just realised I wrote "both" instead of "either" in my original statement and that's not how I meant it, sorry. And what I wished to check was if that is the correct definition of an improper integral.

    I study Riemann integrals, as far as I know. Wikipedia seems to say that statement number 1 (or a corrected version of it) is called "Lebesgue's integrability condition", but I still believe it applies to Riemann integrals right?


    Julien.
     
  8. Apr 12, 2016 #7

    Samy_A

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    Ah yes, that is correct.

    Yes, I think that one is correct.


    That is indeed the definition of an improper integral.
    Yes.
     
  9. Apr 12, 2016 #8
    @Samy_A Thank you for your answers, that was very helpful. I like to be sure I understand definitions well before I go further :)

    Regarding 1 ("bounded functions are integrable, if they are "almost everywhere" continuous"), do you have a suggestion of another criterion to check integrability if the set of discontinuities of the function is uncountable?

    I have another one that says:

    Functions that are "almost everywhere" equal have the same integrals: If f: [a,b] → ℂ is integrable and g: [a,b] → ℂ bounded, and if the set of all x ∈ [a,b] with f(x) ≠ g(x) is countable, then g is also integrable and ∫ab f(x)dx = ∫ab g(x)dx.

    Not sure if that helps in many cases though. What's your opinion about that?

    Julien.
     
  10. Apr 12, 2016 #9

    Samy_A

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    Is that true for Riemann integrals? It is for Lebesgue integral, but I doubt it is for the Riemann integral.
    The Lebesgue integrability condition you mentioned states that for a bounded f: [a,b] → ℂ, f is Riemann integrable if and only if the set of discontinuities of f has Lebesgue measure 0.
    But one can easily take a nice continuous function f, change it's values on the rational numbers in such a way that the resulting function g is nowhere continuous.
    g will be Lebesgue integrable, but not Riemann integrable.
     
  11. Apr 12, 2016 #10
    Well what I quoted in my previous post is absolutely all there is in my script about that criterion. But to be honest, the script my teacher made is very unclear and I doubt it is complete. He has proved in the past to have a certain talent for misleading us :(

    I hear what you're saying with your example, but not even is "Lebesgue" mentioned once in my script, while Riemann is at many occasions. I will ask other students and search a bit on the internet too.

    Thanks a lot again!


    Julien.
     
  12. Apr 12, 2016 #11

    micromass

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    It is if you assume both are integrable. So if ##f## and ##g## are Riemann-integrable and ##f=g## a.e. then the integrals coincide.
     
  13. Apr 12, 2016 #12
    @micromass Hi and thanks for your answer!

    Unfortunately that's not what is stated in my script...It says that if:
    1. f: [a,b] → ℂ is integrable,
    2. g: [a,b] → ℂ is bounded,
    3. the set of all x ∈ [a,b] for which f(x) g(x) is countable,
    then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.

    It does look wrong to me, especially with the example Samy submitted.


    Julien.
     
  14. Apr 12, 2016 #13

    micromass

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    Yes, that is incorrect. It becomes correct for the Lebesgue integral though. It is also correct if we demand finite in (3).
     
  15. Apr 12, 2016 #14
    Alright, thank you for your answer. I reformulate what you said with my own words to make sure I get it right:

    1. f: [a,b] → ℂ is integrable,
    2. g: [a,b] → ℂ is bounded,
    3. the set of all x ∈ [a,b] for which f(x) g(x) is finite,
    then g(x) is integrable and ab f(x)dx = ∫ab g(x)dx.

    About Lebesgue integrals, we didn't do that yet so I guess that criterion for integrability was meant for Riemann integrals.

    Thx a lot,


    Julien.
     
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