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Really stuck: Improper Integrals

  1. Jun 13, 2010 #1
    So I've got my exam of analysis tomorrow, but there's this piece of improper integrals I just don't get... (I'll paraphrase the definition we saw into english)

    So if f is not necessarily Rieman-integrable over ]a,b[ (a and b can be (negative) infinity), but for all c,d with a<c<d<b [tex]\int_c^d f[/tex] exists and the limit of this integral for [tex]c \to a, d \to b[/tex] exists independent of how you approach a and b (relatively to each other), then we call f improperly integrable over ]a,b[.

    Now how can you see that if f is impr. int. over ]a,b[, that it is also impr. int. over ]a,c] and [c,b[ for a certain c in ]a,b[? It was noted in my papers as "evident", but I have no clue of how to prove it (and believe me, I've tried...). The main problem is (with the essence sketched) that with "x + y converges" given, that I have to proof x and y converge seperately. Obviously this is not true in all contextx, but apparently for integrals it is. I know it depends on the fact that in the definition you say the convergence to a limit should be independent of any rows c_n, d_n, but I don't know how to milk that...

    (I see other courses sometimes define and improper integral over ]a,b[ as actually the sum of improper integrals over ]a,c] and [c,b[ and then surely it is evident, but as my course does not take that road, I'd like to be able to show it's equivalent...)

    Thank you very much, I'm clueless.
  2. jcsd
  3. Jun 13, 2010 #2
    In minimum-speak, my problem can be noted as:

    "Given g: ]a,b[ x ]a,b[ -> R and for all sequences (a_n,b_n) that converge to (a,b) is given that g converges to a certain fixed L, proof that in the case of L(x,y) = F(y) - F(x), this implies that for certain constant A,B: F(a_n) -> A, F(b_n) -> B for all sequence a_n and b_n with a_n -> a and b_n -> b"

    But that maybe takes the funk out of things, but I believe that is the essence of my problem.
  4. Jun 13, 2010 #3


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    Essentially that says the the "improper integral"
    [tex]\int_a^b f(x) dx[/tex]
    is given by
    [tex]\lim_{\alpha\to a^+}\int_\alpha^c f(x)dx+ \lim_{\beta\to b^-}\int_c^\beta f(x)dx[/tex]
    where c can be any number strictly between a and b. By the hypothesis "for all c,d with a<c<d<b [tex]\int_c^d f(x)dx[/tex] exists", those integrals exist for all such [itex]\alpha[/itex] and [itex]\beta[/itex] and the hypothesis "limit of this integral for exists" means that limit exists.

    A simple example is the integral [tex]\int_1^\infty \frac{1}{x^2} dx[/tex]. We cannot evaluate the anti-derivative (or any function) at [itex]x= \infty[/itex] but the limit exists.
  5. Jun 13, 2010 #4


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    It looks like it probably is true. Take two separate cases. First let let b_n = b for all n, then the convergence of L(a_n,b) implies the convergence of F(a_n). Similarly let a_n=a for all n to get convergence for F(b_n).
  6. Jun 13, 2010 #5
    Well no, that is what I want to prove, that this follows from the definition in italics in my first post (of course it's evident that it should follow, but if it's truly evident, it should be easy to prove). The statement I quoted from your post is just the thing I want to derive.

    You can't take a constant sequence, as you can see the function isn't defined on a and b (as in the case of improper integrals).

    Thank you both for your time.
  7. Jun 14, 2010 #6

    Bold added.

    I think your problem is simply in making rigorous sense of the first bolded statement above.

    If you do that then you wind up with precisely what HallsofIvy wrote down. So the problem is not really in finding a proof so much as clarifying the definition.

    The only other wrinkle that you may be missing is demonstrating that the expression is independent of your choice of c, but that is pretty easy to show remembering that the function in question is required to be Riemann integrable over any closed subinterval of [a,b].
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