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Close integral

  1. Sep 21, 2009 #1
    what is the theory behind ....if close integral of a expression is zero..then the expression is itself zero
  2. jcsd
  3. Sep 22, 2009 #2


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    Do you mean that if we have an integral whose path is a closed loop that is equal to zero then the integrand must be zero? If that is so then we cannot say that it is true in general, there need to be addtional conditions. For example, the integral of a conservative force over a closed loop will be zero but the integrand, the force, does not necessarily need to be zero itself over the integration path.

    I cannot remember all of the exact conditions where it follows that the integrand has to be identically zero. One example is the Euler-Lagrange equations, in their derivation we can come to the point where,

    [tex] \int_0^1 F(x)\eta(x)dx = 0 [/tex]

    Here, the function \eta(x) is any arbitrary function that enforces the boundary conditions \eta(0)=\eta(1)=0. Since \eta(x) is arbitrary between the endpoints of the path then we conclude that the integrand, F(x), must be identically zero. This is not the exact case that you requested since it is not a closed path but one could extend the same thinking to a closed path.

    That is, if we have an integrand that is the product of two functions, one that is an arbitrary function, then for the integral to be zero the the non-arbitrary function must be zero over the entire integration path. Thus, the integrand itself must be zero.
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