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Help Explaining Integration Trick

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A book I'm reading says
    [tex]\frac{1}{T^2}\int\limits_{0}^{T}\int\limits_{-\zeta}^{T-\zeta}C_x(\tau)d \tau d \zeta = \frac{1}{T}\int\limits_{-T}^{T} \left( 1 - \frac{\left | \tau \right |}{T} \right)C_x(\tau)d \tau[/tex]

    The last expression occurs by reversing the order of integration between [itex]\tau[/itex] and [itex]\zeta[/itex] integration. Now, ..."

    So the part I have no idea how to handle is how to deal with the fact that the tau limits have zeta in them and how that transforms into -T to T (as well as how a |tau| comes out of it too)
  2. jcsd
  3. May 23, 2012 #2
    A good understanding of how to use double integrals to compute volume should help you understand.
  4. May 23, 2012 #3
    Draw the area of integration in the ζτ-plane (if it's easier for you, just replace ζ with x and τ with y).

    HINT: part of the area of integration is in the first quadrant, and another part is in the fourth quadrant.
  5. May 23, 2012 #4
    It's quite simple when you draw the tau-zeta plane. I find the wording in the textbook to be misleading. The order of integration wasn't 'switched'. Entirely new integrals were written that represent the same quantity.

    In case someone finds this thread, here is the solution:

    Tau goes from -zeta to T - zeta while zeta goes from 0 to T. So the bottom limit of tau goes from 0 to -T while the top limit of tau goes from T to 0.

    The equation for the top limit is

    [tex] \tau = -\zeta + T[/tex]

    and for the bottom limit is

    [tex]\tau = -\zeta[/tex]

    If we integrate this area, weighted by C, with zeta first in terms of tau, we visually see if tau > 0, zeta goes from 0 to T - tau. If tau < 0, zeta goes from -tau to T. Since C is independent of zeta, this first integral results simply zeta evaluated at these limits. The integral for tau > 0 becomes T - tau whereas the other becomes T + tau. This can be concisely written at T - |tau|. Finally, seen visually, tau extends from -T to T. So the final answer falls right in place.
  6. May 23, 2012 #5
    Yes it was. Otherwise you couldn't have ended up with a single integral with respect to τ.

    EDIT: I included a graph of the ζτ-plane. From that, we can see that with the order of integration switched, the integration becomes: [tex]\frac{1}{T^{2}}\int^{0}_{-T}\int^{T}_{-τ}C_{x}(τ)dζdτ+\frac{1}{T^{2}}\int^{T}_{0}\int^{T-τ}_{0}C_{x}(τ)dζdτ[/tex]

    From there, you should be able to finish the problem. It seems like you understand the basic idea.

    Attached Files:

    Last edited: May 23, 2012
  7. May 24, 2012 #6
    When I said 'switched', I meant it wasn't only switched, as one would believe with such a blank statement. It was rewritten with two different integrals that were in reverse order. So yes, 'the order of integration' was switched in that zeta went first. However, the beginning integral was not switched -- different ones were. If I were responsible for this manuscript, I would have mentioned different integrals were written with zeta in terms of tau as opposed to tau in terms of zeta, allowing the zeta integration to go first.

    Also, I already got the answer as I previously stated. I'm not sure what about my last post made you think I needed you to rewrite what I already wrote or draw a picture I already drew.
  8. May 24, 2012 #7
    I was trying to be more clear, since you didn't exactly clarify what you meant by "the order of integration wasn't 'switched'". This statement, when interpreteted correctly, is blatantly wrong. You got the right answer obviously since it's already given to you in the problem. But when you said the order of integration wasn't switched, it made me think you might not have fully understood how to arrive at the answer correctly. Hence, why I drew the graph and rewrote the 2 integrals. It didn't occur to me that by "not switched" you meant "rewritten as two integrals with switched order of integration". That just doesn't follow.

    No need to get offended. I was just trying to help. I had a feeling you understood already since the rest of your post made sense for the most part. I was just making sure. A simple "thank you" would be more appropriate.
  9. May 24, 2012 #8
    It wasn't a problem. It was a statement in a textbook on estimation. I am the type that wants to understand all facts in a textbook as opposed to memorizing the result. I am researching the topic for my masters, so understanding is of greater import than when reading a textbook simply to pass a test. I will thank you for your first post. I will also make a point not to thank Whovian whose idea of help is to state vacuous truths. I present a double integral upon which he states I should have no problem comprehending if I have a solid understanding of double integrals.
  10. May 24, 2012 #9
    I greatly respect that. It probably sets you apart from about 90% of the student population. And yes, I agree that the explanation was vague. The author probably assumes that his audience is more interested in the final result than the derivation. So it's good that you searched elsewhere for an explanation.
    Last edited: May 24, 2012
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