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Double Integration change of variables

  1. Feb 29, 2008 #1
    Hi
    I just cannot understand the following transformation, where [tex]\phi(t)[/tex] is the displacement of an optimal path using standard calculus of variations. All functions are defined between 0 and T. [tex]\phi[/tex] equals zero at 0 and T. r is some discount rate, e it the Euler number, t is time.

    [tex]\int^{T}_{0}\theta(y(t))e^{-rt}\int^{t}_{0}\phi(\tau)d\tau]dt[/tex]

    This should be equal to

    [tex]\int^{T}_{0}\int^{T}_{t}\theta(y(\tau))e^{-r\tau}d\tau]\phi(t)]dt[/tex]

    If anybody knows the answer, I would be very happy to get some help here.
    Thanks in advance!
     
  2. jcsd
  3. Feb 29, 2008 #2

    tiny-tim

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    I think you're being blinded by the detail.

    Look carefully and you'll see that the only difference in the two integrands is that t and [tex]\tau[/tex] have been swapped over.

    Now, you can always change the names of the variables. For example, ∫f(x)dx = ∫f(y)dy, or ∫∫f(x,y)dxdy = ∫∫f(z,w)dzdw.

    In your example, x = w and y = z: ∫∫f(x,y)dxdy = ∫∫f(y,x)dydx.

    The only place you have to be careful is in choosing the range of integration.

    It is the triangular area 0 < [tex]\tau[/tex] < t; 0 < t < T.

    In other words, all pairs of t and [tex]\tau[/tex] with [tex]\tau[/tex] < t < T.

    Swapping t and [tex]\tau[/tex] gives: all pairs of t and [tex]\tau[/tex] with t < [tex]\tau[/tex] < T, or:
    :smile: t < [tex]\tau[/tex] < T; 0 < t < T. :smile:
     
  4. Feb 29, 2008 #3
    Thank you so much!!! Your answer makes perfect sense!
     
  5. Feb 29, 2008 #4

    tiny-tim

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    Welcome to PF!

    You're very welcome!
    Which reminds me, I forgot to say …

    :smile: Welcome to PF! :smile:
     
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