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Thread: sin(cLt)*e^(-iLx)

  1. #1
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    sin(cLt)*e^(-iLx)

    I am unsure if this belongs in calculus, differential equation, advance applied math, etc.

    So I would like to figure out the trick to
    $\displaystyle
    g(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\! \sin(c \omega t) e^{-i\omega x} \, d\omega$.
    Yes, I know this is an Inverse Fourier Transform, and I know there are tables. What I am looking for is a good explanation of how
    $\displaystyle
    g(x) = \frac{i}{2}\left[\delta(x+ct) - \delta(x-ct)\right]$.
    Also, I figure this might help me understand the Dirac-delta function a bit better, because I am unsure why
    $\displaystyle f(x) \ast g(x) = f(x + ct) - f(x - ct)$.

    Thank you in advance.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by lvleph View Post
    I am unsure if this belongs in calculus, differential equation, advance applied math, etc.

    So I would like to figure out the trick to
    $\displaystyle
    g(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\! \sin(c \omega t) e^{-i\omega x} \, d\omega$.
    Probably the simplest way to do this is to use the fact that $\displaystyle cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$.

    With that, the integral becomes $\displaystyle g(x)= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega(ct- x)}+ e^{-i\omega(ct+ x}}{2}d\omega$

    Yes, I know this is an Inverse Fourier Transform, and I know there are tables. What I am looking for is a good explanation of how
    $\displaystyle
    g(x) = \frac{i}{2}\left[\delta(x+ct) - \delta(x-ct)\right]$.
    Also, I figure this might help me understand the Dirac-delta function a bit better, because I am unsure why
    $\displaystyle f(x) \ast g(x) = f(x + ct) - f(x - ct)$.

    Thank you in advance.
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  3. #3
    Senior Member
    Joined
    Mar 2009
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    378
    Yeah, I tried this and it was messing me up.
    $\displaystyle
    g(x)= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega(ct- x)} - e^{-i\omega(ct+ x)}}{2i}d\omega
    $
    $\displaystyle
    g(x)= \frac{1}{4\pi i}\left[\frac{-i}{ct - x}e^{i\omega(ct- x)} - \frac{i}{ct + x} e^{-i\omega(ct+ x)} \right|_{-\infty}^{+\infty}
    $.
    This is where I get messed up, and probably where the dirac-delta comes in. If you take these limits they are
    $\displaystyle g(x) = \frac{-1}{4\pi}\left[\frac{+\infty }{ct - x} + \frac{+\infty}{ct + x}\right]$
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