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Homework Help: Confusing Integration Question?

  1. Aug 5, 2012 #1
    Confusing Integration Question??

    1. The problem statement, all variables and given/known data

    The question is as attached.

    3. The attempt at a solution

    I am very confused by the question. Firstly is I(y) a function like (tan-1y) or an operator like ∫y?

    Then they say I(y) - I(∞) = (∏/2) - tan-1y so I suppose that I(y) is basically just -tan-1y.

    I(y) - I(∞) = (∏/2) - tan-1y

    lim(y→0) [ I(y) - I(∞) ] = (∏/2) - 0 = (∏/2)

    But returning to the question, as you let y→0, all you get is

    y(0) = ∫ sinx /x dx = tan-10 = 0

    But that doesn't look right at all.

    I hope someone can clear up all the terminology they are using...

    Attached Files:

  2. jcsd
  3. Aug 5, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Re: Confusing Integration Question??

    The terminology is very clear: I(y) is a function of y, given by a formula in the picture.

  4. Aug 5, 2012 #3
    Re: Confusing Integration Question??

    But when they take the limit of y->0, all you get is I(0) = ∫ sinx /x dx = tan-10 = 0

    When you take the limit of y-> ∞, all you get is I(∞) = 0 (If you substitute ∞ into the question)
    Last edited: Aug 5, 2012
  5. Aug 5, 2012 #4
    Re: Confusing Integration Question??

    They say very plainly that [itex]I(\infty) = 0[/itex], so [itex]I(y) = \frac {\pi} {2} - tan^{-1}(y)[/itex], including [itex]y = 0[/itex].
  6. Aug 5, 2012 #5
    Re: Confusing Integration Question??

    Oh I think I got it now! Cause when you integrate ∂/∂y = - 1/(1+y2) you get

    I(y) = -tan-1y + C

    Based on the first equation, we know y(∞) but not y(0).

    I(∞) = 0, so you solve C = ∏/2...

    Everything makes sense now, thank you!!
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