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Fourier transform of the hyperbolic secant function

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi there!!

    I'm just trying to figure out the Fourier transform of the hyperbolic secant function... I already know the outcome:

    4[itex]\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2) [/itex]

    But sadly, I cannot figure out how to work round to it! :( maybe one of you could help me...

    2. Relevant equations

    I was thinking of using the geometric series 1/(1+q) = [itex]\sum (-q)^n [/itex] for q = e^(-2*x), as the hyperbolic secant is 2*e^(-x)/(1+e^(-2*x)) .
    And then you need to multiply the hyperbolic secant with e^(iωt) and integrate from -∞
    up to ∞.
    But... frankly, that was it. I have never managed to come to the result
    4[itex]\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2) [/itex]

    Maybe someone could give me a hint to to solve this problem...? :)

    Lots of greetings,
    Marie
     
    Last edited: Jan 6, 2013
  2. jcsd
  3. Jan 6, 2013 #2

    Ray Vickson

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    f(x) = sech(x) is an even function, so the FT of f is (g + conjugate g), where ##g = \int_0^{\infty} f(x) e^{i \omega x} \, dx.## Apply this term-by-term to
    [tex] \text{sech}(x) = 2 \sum (-1)^n e^{-(2n+1)x}.[/tex]
     
  4. Jan 6, 2013 #3

    Aaahh, wonderful!! Thank you!

    Just one more question... where does the Term [tex](-1)^n[/tex] come from?

    Thanks a lot!!
    MARIE
     
  5. Jan 6, 2013 #4

    Dick

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    It comes from the expansion of 1/(1+a)=1-a+a^2-a^3+... It's the factor that alternates sign.
     
    Last edited: Jan 6, 2013
  6. Jan 6, 2013 #5
    Oh yes, sure!!! Now I got it :)

    Thank you very much!! :) And have a nice evening!!

    _MARIE_
     
  7. Jan 6, 2013 #6

    Dick

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    You're welcome. Nice evening to you too. Note I goofed on the post. I meant 1/(1+a) not 1/(1-a).
     
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