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Math Help - fourier transform convolutio theorem

  1. #1
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    fourier transform convolutio theorem

    show that  \int_{-\infty}^{\infty}{\frac{\sin(bk)e^{ikx}}{k(a^{2}+k^  {2})}dk}=\frac{\pi \sinh(ab)}{a^{2}}e^{-ax}
    by considering fourier transforms of
     f(x)=e^{-a|x|}  \ \ \ and \ \ \ g(x)=1 when x>b
    Last edited by silversand; April 25th 2009 at 12:36 AM.
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    My advice, is obviously to calculate the fourier transform of these functions, and then use the convolution formula, but in this case use the fact that exp (i b k)=cos(bk)+isin(bk) to replace this with sin(bk) in the integrand.
    if f(k),g(k) are the fourier transforms of f(x),g(x), and h(z) is the convolution of f and g then, then h(x)=F^-1(f(k)g(k)) where F^-1 is the inverse fourier transform.
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    hey

    i know we should use convolution. and i found the transform of f and g.

    but i cant see they can make that expression in the integrand

    anyone can make it more clear?
    Last edited by silversand; April 24th 2009 at 09:12 AM.
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    Quote Originally Posted by silversand View Post
    show that  \int_{-\infty}^{\infty}{\frac{\sin(bk)e^{ikx}}{k(a^{2}+k^  {2})}dk}=\frac{\pi \sinh(ab)}{a^{2}}e^{-ax}
    by considering fourier transforms of
     f(x)=e^{-a|x|}  \ \ \ and \ \ \ g(x)=1
    I think that the function g(x) ought to be a rectangular function. In other words, it is 1 on some interval ([1,1], [1/2,1/2] or [-π,π], depending on which definition of the Fourier transform you are using), and 0 elsewhere. (g obviously can't be 1 on the whole real line because then it would not have a Fourier transform.)

    The integral \int_{-\infty}^{\infty}{\frac{\sin(bk)e^{ikx}}{k(a^{2}+k^  {2})}dk} represents the value at x of (a multiple of) the inverse Fourier transform of the product of two functions, namely \hat{g}(k) = \text{sinc}(bk) := \frac{\sin(bk)}{bk} and \hat{f}(k) = \frac{2a}{a^2+k^2}. But the sinc function is the Fourier transform of a rectangular function, and the function \hat{f} is the Fourier transform of f(x) = e^{-a|x|}. So the inverse Fourier transform of their product will be the convolution of the rectangular function and that exponential function.
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  5. #5
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    Quote Originally Posted by Opalg View Post
    (g obviously can't be 1 on the whole real line because then it would not have a Fourier transform.)
    Not in the conventional sense, but in another existence its FT is \delta(\omega) (give or take a constant multiplier).

    CB
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    que

    So how can you get that  sinh(ab) on the right?
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    If g(x) = \begin{cases}1&\text{if }|x|\leqslant b\\ 0&\text{otherwise}\end{cases} and f(x) = e^{-ax}/{2a} then (g*f)(x) = \int_{-b}^b\frac1{2a}e^{-a(x-t)}dt = \Bigl[\tfrac1{2a^2}e^{-ax}e^{at}\Bigr]_{-b}^b = \frac{e^{-ax}(e^{ab}-e^{-ab})}{2a^2} = \frac{e^{-ax}\sinh(ab)}{a^2}.

    That looks promising, because it gets the right answer apart from a constant multiple. However, I have cheated by using the function e^{-ax} instead of e^{-a|x|}. That doesn't matter provided that x is positive. This means that, in the integral, we want xt to be positive whenever |t|\leqslant b; and that in turn means that I am assuming that x\geqslant|b|. I claim that this is a reasonable assumption. Indeed, if it does not hold then the function \frac{e^{-ax}\sinh(ab)}{a^2} will not have a Fourier transform (in the conventional sense: as the Captain points out, it will have a distributional Fourier transform of some sort), and so this whole method of evaluating the integral will become suspect.
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