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Infinite and improper integral help.

  1. Dec 24, 2004 #1
    I can't seem to evaluate integrals with infinite limits as well as improper integrals. Can anyone help in that? Sorry if this is a little vague but I'm stumped by the whole topic !
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  3. Dec 24, 2004 #2

    matt grime

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    [tex] \int_a^{\infty}f(x)dx[/tex]

    is defined to be the limit as L tends to infinity of

    [tex]\int^L_af(x) dx[/tex]

    when it exists (it may not do).

    [tex]\int_1^{\infty}\frac{1}{x^2}dx = \text{lim}_{L\to \infty}(\frac{-1}{L}+1) = 1[/tex]
  4. Dec 24, 2004 #3


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    To add ot Matt's post,there is no difference between Riemann improper integrals and the ones with infinit limits.The most general form of such an integral is
    [tex] I=\int\int...\int_{R^{n}} f(x_{1},x_{2},...,x_{n}) dx_{1}dx_{2}...dx_{n}=:\lim_{L\rightarrow \infty} \int_{-L}^{+L}\int_{-L}^{+L}...\int_{-L}^{+L} f(x_{1},x_{2},...,x_{n}) dx_{1}dx_{2}...dx_{n} [/tex]

    Sometimes,such integrals diverge.Physics often deals with such integrals and it asks for finite results.
    Many,very many improper integrals are tabulated in the book by the 2 Russian Jews:Gradsteyn & Rytzhik.

  5. Dec 24, 2004 #4

    matt grime

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    That's not correct, must let the lower and upper limits of every integral tend to infinity independently.
  6. Dec 24, 2004 #5


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    It looks like dex is talking about a generalization of the Cauchy principal value of an integral.
  7. Dec 24, 2004 #6
    hmmmm i get the infinite integrals part already. But there are still a few more questions.

    1. How do I find out whether an integral exists or not?

    in perhaps the example,

    [tex] \int_{1}^{\infty} \frac {1}{x+1} dx [/tex],

    why doesn't it exist?

    2. In an improper integral, how do I find the range in which the integral is infinite? Or is it already stated?

    In the example,

    [tex] \int_{0}^{2} (2-x) ^{-1/4} dx [/tex]

    If the substituition of [tex] 2- \delta [/tex] where delta tends towards zero and 0 into the integral, I would get,

    [tex] -\frac {4}{3} \delta ^{3/4} + \frac {4}{3} 2^{3/4}[/tex]

    the final answer would be without the term delta, how do I get rid of it?

    3. Also, when given an integral just like that, how would I be able to tell it apart from an improper integral and just any plain old one?
  8. Dec 24, 2004 #7


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    In you first example, after integrating, we are left with [tex](ln|x+1||_1^L[/tex]. Taking the limit, as L tends to infinity yields infinity, and thus we say that the integral diverges, or does not exist.

    In the second one, the integral is unbounded at x=2, and is thus improper. So [tex] \int_0^2 (2-x) ^{-1/4} dx = \text{lim}_{L\to 2-} \int_0^L (2-x) ^{-1/4} dx [/tex]
    From there, just integrate and take the limit.

    Regarding the third, for an integral to be improper, one of its limits must be infinite, or it must be unbounded at some point over the interval.

    Then again, I may have completely misunderstood your questions.
  9. Dec 26, 2004 #8
    Hey I've finally understood it, thanks. But there is still the part where,......

    How do I know whether an integral is unbounded at a certain interval? Is it usually because of fractions where there is 1/0 and this yields infinity? Is there any other cases?
  10. Dec 27, 2004 #9


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    Yes, it is probably most often when you have a function of the form 1/x but in general, the function is unbounded at x=n when [tex] \text{lim}_{x\to n} f(x) =\infty [/tex]

    Thats just a rule off the top of my head so that may be wrong. Notable cases that come to mind other than 1/x are both log and ln, which are unbounded for x=0. Hope that helps.
    Last edited: Dec 27, 2004
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