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Integration by parts and coefficients

  1. Sep 16, 2008 #1
    [tex] b_{n} = \frac{1}{\pi}\int^{\pi}_{-\pi}sin\theta sin n\theta d \theta [/tex]

    let

    [tex] u = sin \theta, \ du = cos \theta d \theta [/tex]
    [tex] dv = sin n \theta d \theta, \ v = -\frac{1}{n}cosn \theta [/tex]

    [tex] = \left[-\frac{1}{n} sin \theta cos n \theta \right|^{\pi}_{-\pi} + \frac{1}{n} \int^{\pi}_{-\pi} cos \theta cos n \theta d \theta \right] [/tex]

    now [tex]-\frac{1}{n} sin \theta cos n \theta \right|^{\pi}_{-\pi} = 0 [/tex]

    and

    [tex] u = cos\theta, \ du = -sin \theta d\theta [/tex]

    [tex] dv = -sin\theta d\theta, \ v = \frac{1}{n}sin n \theta [/tex]

    [tex] = \frac{1}{n} \left[\frac{1}{n} cos \theta sin n \theta \right|^{\pi}_{-\pi} + \frac{1}{n} \int^{\pi}_{-\pi} sin \theta sin n\theta \right] [/tex]

    I keep getting that to come out to zero but I know it shouldn't. I'm not sure what it should come out to, but it's a Fourier coeff for expanding sin(x). Since sin(x) is an odd function, I know that this coeff should have a value other than 0.

    Can someone help me?
     
  2. jcsd
  3. Sep 16, 2008 #2
    You have to do "integration by parts" twice to get the answer. You are correct - it's not zero.
     
  4. Sep 16, 2008 #3
    well I did, and got zero. The second integration by parts is the last line. What am I missing here?
     
  5. Sep 16, 2008 #4
    I don't have time to go through your entire calculation, but one of the denominators should be n-squared and not n. Then the terms don't cancel. I hope this helps.
     
  6. Sep 16, 2008 #5
    Ok thanks for looking. I'll keep trying.
     
  7. Sep 16, 2008 #6
    [tex] b_{n} = \frac{1}{\pi}\int^{\pi}_{-\pi}sin\theta sin n\theta d \theta [/tex]

    let

    [tex] u = sin \theta, \ du = cos \theta d \theta [/tex]
    [tex] dv = sin n \theta d \theta, \ v = -\frac{1}{n}cosn \theta [/tex]

    [tex] = \left[-\frac{1}{n} sin \theta cos n \theta \right]^{\pi}_{-\pi} + \left[\frac{1}{n} \int^{\pi}_{-\pi} cos \theta cos n \theta d \theta \right] [/tex]

    now [tex]\left[-\frac{1}{n} sin \theta cos n \theta \right]^{\pi}_{-\pi} = 0 [/tex]

    and

    [tex] u = cos\theta, \ du = -sin \theta d\theta [/tex]

    [tex] dv = -sin\theta d\theta, \ v = \frac{1}{n}sin n \theta [/tex]

    [tex] = \frac{1}{n} \left[\frac{1}{n} cos \theta sin n \theta \right]^{\pi}_{-\pi} + \frac{1}{n} \left[\frac{1}{n} \int^{\pi}_{-\pi} sin \theta sin n\theta \right] [/tex]

    since [tex]\frac{1}{n} \left[\frac{1}{n}cos\theta sin n \theta \right]^{\pi}_{-\pi} = 0 [/tex]

    Then

    [tex] \frac{1}{\pi}\int^{\pi}_{-\pi}sin\theta sin n\theta d \theta = \frac{1}{n^{2}}\int^{\pi}_{-\pi}sin\theta sin n\theta d \theta [/tex]

    That still looks like zero to me. ??? :(
     
  8. Sep 17, 2008 #7
    Sorry - yesterday I never actually evaluated my answer.

    After reconsidering the problem, I realize that because the integrand is the sine of theta - an odd function - times the sine of (n times theta) - another odd function - the integrand is an even function. Therefore integrating over limits that are symmetrical to theta = zero will yield an answer of zero.

    I verified this by evaluating my answer from negative pi to pi.

    So, I now believe your original answer is correct, despite your misgivings.
     
  9. Sep 17, 2008 #8
    Weird. Shouldn't be zero. I was trying to find the coefficients for a Fourier series exp. of sin(theta). All the an's should be zero, but the bn's should be some value.
     
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