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Fourier Transform Proof

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    If F(p) and G(p) are the Fourier transforms of f(x) and g(x) respectively, show that

    ∫f(x)g*(x)dx = ∫ F(p)G*(p)dp

    where * indicates a complex conjugate. (The integrals are from -∞ to ∞)

    2. Relevant equations

    F(p) = ∫f(x)exp[2∏ipx]dx
    G(p) = ∫g(x)exp[2∏ipx]dx
    G*(p) = ∫g(x)exp[-2∏ipx]dx
    f(x) = ∫F(p)exp[-2∏ipx]dp
    g(x) = ∫G(p)exp[-2∏ipx]dp
    g*(x) = ∫G(p)exp[2∏ipx]dp

    3. The attempt at a solution

    Well this question is kind of weird to me since most of the in class examples have been based on knowing the function and then using different methods of integration to find the transforms, but in this proof it's all arbitrary, obviously.

    Well simply subbing definitions in I get:

    ∫f(x)g*(x)dx = ∫[∫F(P)exp[-2∏ipx]dp∫G(p)exp[2∏ipx]dp]dx


    ∫F(p)G*(p)dp = ∫[∫f(x)exp[2∏ipx]dx∫g(x)exp[-2∏ipx]dx]dp

    Now I guess if I can show that these two lines simplify to the same thing I have my proof. However, I am not sure how to simplify this. Maybe I am forgetting some basic property of integrals?

    Also I have no idea if this approach is even correct and it would be better to start someplace else.
  2. jcsd
  3. Feb 13, 2012 #2


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    You need to be a bit more careful with the integration variables. You have
    f(x) &= \int F(p)e^{-2\pi ipx}\,dp \\
    g^*(x) &= \int G^*(p')e^{2\pi ip'x}\,dp'.
    \end{align*} Remember that F(p) and G(p) are complex. So now you have
    \int f(x)g^*(x)\,dx &= \int\int F(p)e^{-2\pi ipx}\,dp \int G^*(p')e^{2\pi ip'x}\,dp' \,dx \\
    &= \iiint F(p) G^*(p') e^{-2\pi ipx} e^{2\pi ip'x}\,dp\,dp'\,dx
    \end{align*} Now you want to identify the Dirac delta function that's in there somewhere.
  4. Feb 13, 2012 #3
    Do I have my signs backwards? When taking the fourier transform on a function do I have a negative exponential? My class notes don't match will other things I've found...
  5. Feb 13, 2012 #4


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    It depends on what convention your instructor has decided to use. There can also be factors of ##2\pi## that move around depending on the convention you choose.

    The Wikipedia entry on the Fourier transform lists the common ones.
  6. Feb 13, 2012 #5
    My instructor has the 2∏ in the exponential. Yes I've seen cases where it is infront of the integral as a 1/(2∏). I don't fully understand the (non) difference, just sticking with what I've seen.
  7. Feb 13, 2012 #6
    Aha, I believe I have to use

    FT of g*(x) = G*(-x)
    Fixes my sign issue I believe.
  8. Feb 13, 2012 #7
    Ugh, still have 1 sign error tripping me up...

    going by my definitions..

    replacing only g*(x)

    ∫f(x)g*(x)dx = ∫f(x)[∫G(p)exp[2∏ipx]dp]dx = ∫f(x)[∫G*(p)exp[-2∏ipx]dp]dx

    = ∫G*(p)[∫f(x)exp[-2∏ipx]dx]dp

    In that last line if the negative wasn't there I could replace it with F(P) from y definitions and be done....
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