1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Fourier Integration
  1. Fourier integrals (Replies: 1)

  2. Fourier Integral (Replies: 5)

  3. Fourier integral (Replies: 2)