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Help with integral question.

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Here is the question given:

    jp09x.png


    2. Relevant equations



    3. The attempt at a solution

    SO i used the product to sum angle formula:
    sin u sin v = 1/2[cos(u-v) - cos (u + v)]

    so I get
    from ∏ to -∏

    1/2 ∫cos(k-n)x dx - ∫cos(k+n)x dx

    = 1/2 [1/k-n (sin (k-n)∏ - sin(k-n)-∏) -1/k+n(sin(k+n)∏ - sin(k+n)-∏)]

    so when k isn't equal to ±n, i think the second term becomes 0, but the first term doesn't have to be 0 as well (for the first condition).
     
  2. jcsd
  3. Jan 19, 2012 #2

    micromass

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    What is [itex]\sin(n\pi)[/itex] if [itex]n\in \mathbb{Z}[/itex]???


    Also note that your method does not work if n=k. Indeed, you seem to have a term [itex]\frac{1}{n-k}[/itex] there. But if n=k, then you will divide by zero. So if n=k, then you'll need to do something different.
     
  4. Jan 19, 2012 #3

    SammyS

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    sin(mπ) = 0 for all integers, m. So when k≠n, all those terms are zero.

    When k=n, the last term is zero. The first term must be evaluated as a limit n → k .
     
  5. Jan 19, 2012 #4
    Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then
     
  6. Jan 19, 2012 #5

    SammyS

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    Actually it's of the form 0/0 .
     
  7. Jan 19, 2012 #6

    micromass

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    If k=n, then

    [tex]\int_{-\pi}^\pi \cos(n-k)x dx=\int_{-\pi}^\pi \cos(0)dx=\int_{-\pi}^\pi dx[/tex]
     
  8. Jan 19, 2012 #7

    SammyS

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    If n=k, the integrand becomes sin2(kx).
     
  9. Jan 19, 2012 #8
    I see what you did but k+n doesn't necessarily have to be an integer does it? What if k+n= 0.5, it doesn't make it 0.
     
  10. Jan 19, 2012 #9

    micromass

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    I think the question implies that both n and k are integers. If n and k are not integers, then the result is not true.
     
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