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Fourier transform of sin(x)

  1. Nov 17, 2009 #1
    I apologise in advance, I don't know latex or any of those math-prettifiers, so this'll be fairly crude. I'll opt to write thing in words to avoid confuson where possible...

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

    Find the exponetial fourier transform of the given f(x) and write f(x) as a fourier integral.

    f(x) = sin(x), for |x|< pi/2, 0 for |x|> pi/2

    2. Relevant equations

    f(x)= integral from negative infinity to infinity of g(alpha)*e^(i*alpha*x) dalpha
    g(alpha)= 1/(2*pi) * integral from negative infinity to infinity of f(x) e^(-i*alpha*x) dx

    3. The attempt at a solution

    Using the above formula for g(alpha) and from a table of integrals the fact that integral sin(u)*e^(a*u)du = (e^(au))/((1+a^2))*(a*sin(u)-cos(u)), I eventually arrived at:

    g(alpha) = (alpha*cos(alpha*pi/2))/(i*pi*(1-alpha^2))

    Which can simply be stuck into the integral formula above to get the fourier integral.

    I don't have much experiance with these though, and that seems like it might have an extra i term, or alpha term or something. Could someone check it for me?

    I know that the same problem replaced with cos instead of sin gives a final result of
    g(alpha)=cos(alpha*pi/2)/(pi*(1-alpha^2)), so the same thing without an 1/i or alpha coefficient. The i term feels right (since based on eulers formula, i goes with sin), but I dunno about tht extra alpha term I got....
     
  2. jcsd
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