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N-th integral

  1. Nov 17, 2013 #1
    If exist differentiation until the nth order, so, why "no exist" integration until the nth order too? I never saw a quadruple or quintuple integral, and if exist, it's always with respect to different variables. Why not difine an integral so?

    [tex]\int\int\int f(x)dx^3[/tex]
     
  2. jcsd
  3. Nov 17, 2013 #2

    mfb

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    It is possible, I just don't know of any applications.
     
  4. Nov 17, 2013 #3

    arildno

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    Why bother?
    Even the first integral is non-unique, what point would it be to gain a long, polynomial tail from your subsequent antidifferentiations??
     
  5. Nov 17, 2013 #4
  6. Nov 17, 2013 #5

    Mark44

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    I've never seen one written this way; i.e., with dx3. The usual way things are done is to have different variables of integration, something like this:
    $$\int_a^b \int_c^d \int_e^f f(x, y, z) dz dy dx$$

    or even like this:
    $$\int_a^b \int_c^d ~...~\int_e^f f(x_1, x_2, ..., x_n) dx_1~ dx_2~...~ dx_n$$
    Here we're integrating over a subset of Rn.
     
  7. Nov 17, 2013 #6

    Curious3141

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    It is possible to define a repeated integral, although the notation used is usually ##dx...dx## (n times) rather than ##dx^n##. The nth such repeated integral can be denoted ##f^{-n}(0)##.

    And as long as the constants of integration at every step can be justifiably "ignored", e.g. ##f^{-1}(0) = ... = f^{-n}(0) = 0##, then it's easy to derive and prove a simple formula for the general repeated integral. See: http://mathworld.wolfram.com/RepeatedIntegral.html

    You can derive it with integration by parts and prove the form by induction. Wiki also has something on this: http://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

    The general formula has an application in defining fractional integration.
     
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