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Question on integration bounds

  1. Feb 18, 2013 #1

    joshmccraney

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    hey everyone

    is integrating from [itex]-L<x<L[/itex] the same as integrating over [itex]-L \leq x \leq L[/itex]? im looking for a rigorous response, so if you have time could you explain why, rather than simply yes or no?

    thanks!
     
  2. jcsd
  3. Feb 18, 2013 #2

    Zondrina

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    If you're speaking about a Riemman Integral, then last I checked there was zero contribution from the endpoints.
     
  4. Feb 18, 2013 #3

    pwsnafu

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    Which definition of integration are you using?
     
  5. Feb 18, 2013 #4

    jbunniii

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    Assuming the function is integrable on ##[-L,L]##, it will have the same integral on ##(-L,L)##.

    For the Lebesgue integral, the reason is simple: the two endpoints have measure zero, so they do not contribute to the value of the integral.

    The Riemann integral is only defined on closed intervals, so we have to define what is meant by integrating over ##(-L,L)##. The usual way to do this is to use improper integrals:
    $$\lim_{\epsilon \rightarrow 0^+} \int_{-L + \epsilon}^0 f(x) dx + \lim_{\epsilon \rightarrow 0^+} \int_0^{L - \epsilon} f(x) dx$$
    Note that we have
    $$\int_{-L}^{L} f(x) dx = \int_{-L}^{-L + \epsilon} f(x) dx + \int_{-L + \epsilon}^0 f(x) dx + \int_0^{L - \epsilon} f(x) dx + \int_{L - \epsilon}^{L} f(x) dx$$
    so to prove what we want, we simply need to show that
    $$\lim_{\epsilon \rightarrow 0^+} \int_{-L}^{-L + \epsilon} f(x) dx = \lim_{\epsilon \rightarrow 0^+} \int_{L - \epsilon}^{L} f(x) dx = 0$$
    But this is quite straightforward; since ##f## is integrable over ##[-L,L]##, by definition it is bounded on that interval, say ##|f(x)| \leq M## for all ##x \in [-L,L]##. Then
    $$\left| \int_{-L}^{-L + \epsilon} f(x) dx \right| \leq \int_{-L}^{-L + \epsilon} |f(x)| dx \leq \int_{-L}^{-L + \epsilon} M dx = M\epsilon$$
    which converges to 0 as ##\epsilon \rightarrow 0^+##. We can do the same thing for the other small interval.
     
  6. Feb 18, 2013 #5
    pwsnafu, I'm curious what integral definitions produce different answers in this case?
     
  7. Feb 18, 2013 #6

    pwsnafu

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    If ##\mu## is a measure on ℝ, then the integral wrt ##\mu## gives
    ##\int_{[a,b]} f \, d\mu = \int_{(a,b)} f \, d\mu + f(a)\, \mu(\{a\}) + f(b) \, \mu(\{b\})##.

    So for any atomless measure (including the Lebesgue measure) the integrals are equal.
    But if ##\mu(\{a\}) \neq 0## then the integrals are different.
     
    Last edited: Feb 18, 2013
  8. Feb 18, 2013 #7

    joshmccraney

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    and for [itex]f(x)[/itex] to be integrable on ##[-L,L]##, is it sufficient [itex]f(x)[/itex] need only exist on that interval provided only a finite amount of discontinuities exist? i ask because i am integrating over a piecewise-smooth function.

    for the record i am talking about a reimann integral. apologies for the ambiguity.

    thanks i appreciate your response(s)!
     
  9. Feb 18, 2013 #8

    pwsnafu

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    A function is Riemann integrable if the set of discontinuities has measure zero. In particular if there are a finite number of jump discontinuities, you have no problem.
     
  10. Feb 18, 2013 #9

    jbunniii

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    Just for the sake of completeness I'll mention that the function must be bounded, and that the condition is both necessary and sufficient: a bounded function on a finite length closed interval [a,b] is Riemann integrable if and only if the set of discontinuities has measure zero.
     
  11. Feb 19, 2013 #10

    jbunniii

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    By the way, another way to see this is to observe that if ##f## is integrable over ##[-L,L]##, then integrating ##f## over ##(-L,L)## is equivalent to setting ##f(L) = f(-L) = 0##, and integrating over ##[-L, L]##. So the problem reduces to showing that if you change the value of a function at a finite number of points, the integral does not change. It's easy to see that this in turn is equivalent to showing that if you integrate a function which is zero everywhere except at a finite number of points (it suffices to consider just one point, by linearity), the result is zero. This is an easy Riemann sum argument.
     
  12. Feb 19, 2013 #11

    joshmccraney

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    thanks. i suddenly feel like an idiot for not realizing this.
     
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