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Complex Fourier Transform & Its Inverse (also Dcontour integrals)

  1. Dec 6, 2005 #1
    For the function [tex]f(x)[/tex] given by:
    [tex]f(x) = e^{2x} (x<0), = e^{-x} (x>0)[/tex]
    I have got the complex Fourier Transform to be:
    [tex] F(k) = {3(k^{2} + ik + 2)}/{(k^{2}+1)(k^{2} + 4)}[/tex]
    Now I'm trying to verify the formula for the inverse transform by using a D-contour integral. Just taking the x>0 case I have found the strip of regularity and am closing the D- contour below (to avoid the exponential exploding).
    Closing the contour below gives 3 poles in the contour, namely:
    I have "argued away" the curve of the Dcontour okay.
    So now computing the sum of the residues and multiplying by (-2pi i) should give me back:
    [tex]f(x) = e^{-x}[/tex]
    The residues at k= i and k= -2i are zero so just working on k= -i:
    [tex]res = {3(k^{2}+ik+2)e^{-ikx}}/{4k^{3} + 10k}[/tex]
    I got this by just differentiating bottom line (trick for getting formula for res)
    when k = -i this gives:
    [tex]res = {-2e^{-x}}/(i)[/tex]
    multiplying by (-2pi i) and dividing by 2pi (according to inv transform formula) gives..
    [tex]f(x) = 2e^{-x}[/tex]
    why have I got that 2?!
    Last edited: Dec 6, 2005
  2. jcsd
  3. Dec 7, 2005 #2


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    If you back up a step, your transform is

    [tex]F(k) = \frac {3}{(k + i ) (k - 2i)}[/tex]

    which gives the correct inverse transform so you have made an algebraic error using the "rationalized" form.

    Incidentally, when you close the contour in the lower half plane you don't encounter the pole at k = i at all.
  4. Dec 7, 2005 #3
    Thanks Tide.

    I made an error in squaring -i! All that work and I just couldn't see my mistake at all.

    Sleeping on it helped alot!!

    Thanks again.
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