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Quick question about integration

  1. Sep 13, 2005 #1
    I have a quick question about integration after a change of variables has been made.

    Suppose there is a function [tex] R(t_{1},t_{2}) [/tex] that actually just
    depends on the difference [tex] t_{1} - t_{2} [/tex]. The goal is then to
    simplify the following integral:

    \frac{1}{T^{2}}\int^{T}_{0}\int^{T}_{0} R(t_{1},t_{2}) dt_{1}dt_{2}

    by using the substitution [tex] t_{1}' = t_{1} [/tex] and [tex] t_{2}'= t_{1} - \tau [/tex].

    A straight substitution yields:
    \frac{1}{T^{2}}\int \int^{T}_{0} R(\tau) dt_{1}'(dt_{1}' - d\tau)

    I am uncertain about two things:

    1) the integration bounds on the outer integral after the substitution has been made
    2) whether or not [tex]dt_{1}'[/tex] in the outer integral is zero since
    [tex]dt_{1}[/tex] is held constant when integrating over [tex]dt_{2}[/tex] before the substitution was made.

    As a heads up the final result is supposed to be:
    \frac{1}{T^{2}}\int^{T}_{-T}\left(T-\left|\tau\right|\right)R(\tau) d\tau

    Thanks in advance for any responses.
  2. jcsd
  3. Sep 13, 2005 #2


    User Avatar
    Science Advisor

    To simplify writing I will use s instead of tau.

    Let s=t1-t2, t=t2. Then the differential looks like R(s)dsdt, with s limits -t,T-t and the t limits 0,T.

    Next reverse the order of integration. The t limit is split into 2 parts. For s<0, the limits are -s and T, while for s>0, the limits are 0 and T-s. The s limits are -T and T.

    When you integrate with respect to t, you get T-|s|.
  4. Sep 13, 2005 #3
    ok I see it now, Thanks. Also, the determinant of the Jacobian of the coordinate transformation is 1, so I guess it is correct to say that dt1dt2=dsdt.
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