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Homework Help: Fourier Integration

  1. Jan 5, 2009 #1
    I am studying Fourier for an exam and came across something in my notes that I can't get my head round, might be a simple integration issue. Let me explain.

    1. The problem statement, all variables and given/known data
    The tutorial question in my notes that I am studying is as following:

    1. Consider the periodic function defined by f(t) = {[tex]\frac{-1 \ \ \ \ -\pi \leq t \leq 0}{1 \ \ \ \ 0 < t < \pi}[/tex]
    Find its Fourier expansion.



    2. Relevant equations
    a0 = 0 (because odd function)
    an = 0 (because odd function)
    bn = [tex]\frac{2}{\pi} \int^{\pi}_{0} f(t) \ sin \ nt \ dt[/tex]


    3. The Solution written on my notes:
    bn = [tex]\frac{2}{\pi} \int^{\pi}_{0} 1 \ sin \ nt \ dt[/tex]
    bn = [tex]\frac{2}{\pi} \left[-\frac{1}{n} \ cos \ nt\right]^{\pi}_{0}[/tex]


    My question is, how can you get [tex]-\frac{1}{n} \ cos \ nt [/tex] when integrating [tex] 1 \ sin \ nt[/tex].

    Should it not have been [tex]t \ cos \ nt[/tex] if integrating with respect to t (dt)?

    I know it might be a simple answer but I have been studying for a while now and can't get my head round this, are my notes incorrect?

    Note: I have it worked out in my notes down to the solution where [tex] f(t) = \frac{4}{\pi}(sint+\frac{1}{3}sin3t+\frac{1}{5}sin5t+\frac{1}{7}sin7t+...)[/tex] I have omitted most of the working out and most of my notes as they are irrelevant to my question.
     
  2. jcsd
  3. Jan 5, 2009 #2

    Dick

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    To integrate 1*sin(nt)=sin(nt) you just substitute u=n*t, du=n*dt. You don't get t*cos(nt). That's just wrong. Try differentiating t*cos(nt) (use the product and chain rules). You don't get sin(nt).
     
  4. Jan 5, 2009 #3

    HallsofIvy

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    Surely you know better than that! The integral of sin(nt) is (-1/n)cos(nt) because the derivative of cos(nt) is - n sin(nt). I have no idea why you would want to multiply by "t"!

     
  5. Jan 5, 2009 #4
    Right, I understand Fourier no problem, its the stupid simple integration that is giving me headaches :(

    if it was [tex]\int 1 + sin(nt)[/tex] then you would get [tex]\left[t + \frac{1}{n} cos nt \right][/tex] Wouldn't you?
     
  6. Jan 5, 2009 #5

    Dick

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    You would get t-(1/n)*cos(nt). There's a sign problem. But there's no product rule for integration. Integrating 1*sin(nt) has nothing to do with integrating 1.
     
  7. Jan 5, 2009 #6
    Right, sorted cheers!!
     
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