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Passing limits through integrals

  1. Mar 30, 2009 #1
    This seems to be an elementary question, but I rarely have to deal with rigor these days, so excuse me if there's a simple answer.

    I'm looking for a theorem that will allow me to state,

    F(z) = \int_0^b \frac{f(\tau)}{\tau - z} d\tau \sim \int_0^b \frac{f(\tau)}{\tau} d\tau

    as [itex]z \to 0[/itex] and [itex]f(\tau): \mathbb{C} \to \mathbb{C}[/itex], but it's real along the path of integration.

    I believe [itex]f(\tau)[/tex] satisfies these properties:
    • [itex]f(0) = 0[/itex] and [itex]f(b) = 0[/itex]
    • [itex]f(\tau)[/itex] is Holder continuous along the path of integration

    Note that F(z) is well-defined and Holder continuous along the line because of the values of f at the endpoints.

    An example would be,

    \int_0^1 \frac{\sqrt{\tau}}{\tau - z} d\tau
  2. jcsd
  3. Mar 30, 2009 #2


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    What is it that you are looking for. Do you want a theorem that will tell you that the integral you wrote down is a primitive of f(z), or do you want a theorem that will tell you that you can replace t - z by t if z goes to zero.
    Because the latter simply follows from continuity of F(z) at z = 0.
  4. Mar 30, 2009 #3
    Huh. I guess it's that simple. Sorry, tripped over myself there.
  5. Mar 30, 2009 #4


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    Would it make you feel less stupid to know that I usually look such things up, like when I really need to know the conditions of passing limits through integrals? Otherwise I usually write something like: "I don't know the exact theorem, but f is continuously differentiable which is - although overkill - definitely a sufficient condition." :smile:
  6. Apr 2, 2009 #5
    Your example doesn't have f(b)=0...whuuuu?
    How is f(x) Holder continuous along the real line segment [0,1]?
    The derivative of the square root function is unbounded as x-> 0.

    A basic convergence thm I would consider is the Lebesgue Dominated Convergence Thm.

    Tough going to just deem something continuous, quote a thm and call it a day.
    You have to have gone to a high-tuition college to get away with that and still you
    may be wrong.
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