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Fourier Series Integral

  1. Nov 9, 2011 #1
    Hey there,

    First time user of this forum.

    I have a question regarding an integral i've been stuck on for the past few days. I would really appreciate any eye opener into this problem!

    How do I find the Fourier coefficients for f(x)=xcos(x), x [-∏,∏]

    So when calculating the coefficient bn I figure that the equation I'll be working with is (1/∏)∫xcos(x)sin(kx)dx (FROM -∏ to ∏).

    I keep getting down to 1/2∏∫xsin((k+1)x)dx (FROM -∏ to ∏) + 1/2∏∫xsin((k-1)x)dx (FROM -∏ to ∏) . I some how always get a 0 for b1. What am I doing incorrectly??
  2. jcsd
  3. Nov 9, 2011 #2

    I like Serena

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    Welcome to PF, tigertan! :smile:

    I'm not getting 0 for b1.
    So what did you do?
  4. Nov 10, 2011 #3
    Am I right in breaking it up to 1/2∏∫xsin((k+1)x)dx (FROM -∏ to ∏) + 1/2∏∫xsin((k-1)x)dx (FROM -∏ to ∏)?

    I then end up with 1/2∏[-(k+1)cos((k+1)x) + sin((k+1)x)](FROM -∏ to ∏) + 1/2∏[-(k-1)cos((k-1)x) + sin((k-1)x)](FROM -∏ to ∏)

    This equals to 0 when I put b1
  5. Nov 10, 2011 #4

    I like Serena

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    You lost a factor x in your integration.
    It seems you did not properly apply integration by parts.

    You should have an expression containing: x cos(k+1)x.
    When you substitute the bounds -pi and +pi, these add up.
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