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Fourier transform of integral

  1. Jun 17, 2011 #1
    Hey, this is my first post, great forum!! You've really helped me a lot of times.
    I have a problem solving an integro-differential equation. It involves a term of the form: integration over [t, +infinity) of f(s)*exp(t-s)ds.
    I have to solve the equation using Fourier transform, and most of the other terms are ok, but I have a problem with this one. I know that if the limits of integration where (-infinity, +infinity) it would be a convolution and it's fourier transform would be simple, but I don't know what to do in this case. I tried to find the Fourier transform using the logic of the proof of the convolution theorem, but I end up with a result with t and s. Do you have any ideas on what to do? Any ideas and help would be really welcome!
  2. jcsd
  3. Jun 17, 2011 #2


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    Hi lost87 and welcome to Physics Forums.

    Could you perhaps post the full integro-differential equation so that we can take a look at it?
  4. Jun 17, 2011 #3
    So, what you are describing is like convolution, only integrating over s>=t. I think there is way you could restate that integral as an equivalent integral from negative infinity to infinitiy by using a different function in place of f(s).
  5. Jun 17, 2011 #4
    Hey, the equation has the form: f'=af(t)+b(integral of f(s)*exp(t-s) ds from t to +infinity) +cf''+e(integral of f(s)*exp(-(t-s)) ds from -infinty to t). I can find the Fourier transform for most of the terms, but I don't know what to do with the terms with the integral. I hope this helps.
  6. Jun 17, 2011 #5
    Hey, thanks for your reply, but I am not sure that I understand what you are suggesting. Will that involve a change of the variables s, t?
  7. Jun 17, 2011 #6
    No, I was suggesting replacing f(s) with a different function of s, so that the bounds of the integral could be changed to from -infinity to infinity, and the integral would be equal to the integral you described.

    Is this a homework problem?
    Last edited: Jun 17, 2011
  8. Jun 18, 2011 #7
    Hey, thanks again for your reply. No it's not a homework problem, it's just something I am working on in my dissertation
  9. Jun 18, 2011 #8
    u(s-t) evaluates to zero when s < t

    [itex]\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds + \int^{\infty}_{t}f(s)u(s-t)e^{t-s} = \int^{\infty}_{t}f(s)e^{t-s}ds[/itex]
    [itex]\int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds = 0 [/itex]

    Now, to get the Fourier transform of [itex]\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds [/itex], we have to figure out what two functions [of t] are being convolved. I hadn't thought this all the way through so I may have been misleading when I suggested replacing f(s). Actually, we'll have to group the unit step with the exponential to use the convolution theorem.

    [itex]g(\tau) = u(-\tau)e^{\tau}[/itex]
    so that
    [itex]g(t-s) = u(s-t)e^{t-s}[/itex]

    [itex]\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{\infty}_{-\infty}f(s)g(t-s)ds[/itex]


    [itex]\int^{\infty}_{-\infty}( \int^{\infty}_{t}f(s)e^{t-s}ds)e^{-j\omega t}dt = \mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \mathcal{F}\left\{ f(t) * g(t) \right\}[/itex]

    [itex] = \mathcal{F}\left\{ f(t) \right\}\mathcal{F}\left\{ g(t) \right\} = \mathcal{F}\left\{ f(t) \right\} \mathcal{F}\left\{ u(-t)e^{t} \right\}[/itex]

    [itex]\mathcal{F}\left\{ u(t)e^{-t} \right\} = \frac{1}{1 + j\omega}[/itex]
    [itex]\mathcal{F}\left\{ u(-t)e^{t} \right\} = \frac{1}{1 - j\omega}[/itex]


    [itex]\mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \frac{\mathcal{F}\left\{ f(t) \right\} }{1 - j\omega}[/itex]
  10. Jun 19, 2011 #9
    This is great! I can't explain how helpful this has been! Thanks for everything!
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