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Difficult integrals (important für Fourier series)

  1. Nov 2, 2008 #1
    Hello,

    I couldn't figure out the following problem so far.

    1. The problem statement, all variables and given/known data

    [tex]m,n \in \mathbb{N}[/tex]. Show the following relations:
    [​IMG]

    2. Relevant equations

    a) is no problem, but b) and c) seem tricky. I tried partial integration (two times) on b), which gave me a term which contained the original integral. So I sorted the term and got that:

    [tex]\int_{0}^{2\pi}\sin\left(mx\right)\cos\left(nx\right)dx=\frac{1}{1+\frac{n^{2}}{m^{2}}}\left(\left[-\frac{\cos\left(mx\right)}{m}\cos\left(nx\right)\right]_{0}^{2\pi}+\frac{n}{m}\left[\frac{\sin\left(mx\right)}{m}\sin\left(nx\right)\right]_{0}^{2\pi}\right)[/tex]

    But I can't see why this should be zero.

    Any help appreciated!

    kP
     
  2. jcsd
  3. Nov 2, 2008 #2

    Redbelly98

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    Welcome to PF.

    I think the trig identities for angle addition and subtraction would be useful here. I mean the ones that involve
    sin(a+b), sin(a-b), cos(a+b), and cos(a-b)

    EDIT:
    I looked more carefully at what you did get. Okay, if m and n are integers, what can you say about cos(mx) and cos(nx) when they are evaluated at 0 vs. when they are evaluated at 2pi?
     
    Last edited: Nov 2, 2008
  4. Nov 2, 2008 #3
    Thank you Redbelly,

    you are right, I don't know why I missed that, n and m being integers, b) is quite obviously 0.

    Concerning c): Yes, I also thought about those trig identities, but still couldn't figure it out. It's

    cos(mx+nx)+sin(mx)sin(nx)=cos(mx)cos(nx), but trying to integrate the LHS, I get into even more trouble.
     
  5. Nov 2, 2008 #4

    Redbelly98

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    Try

    cos(mx+nx) + cos(mx-nx) = ?
     
  6. Nov 2, 2008 #5
    That's it, thanks a lot. Works fine! :-)

    EDIT: BTW, if an admin reads that, please change the thread title from the German "für" to for ;-)
     
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