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Dirac delta integration

  1. Jul 11, 2011 #1
    1. The problem statement, all variables and given/known data


    [tex]\int_{-\infty}^{+\infty} x(t) \delta (\beta t - t_{0}) dt [/tex]

    for [tex] x(t) = e^{a t} u(t) [/tex]

    there is no information conserning a, β, or [tex]t_{0}[/tex]...

    3. The attempt at a solution

    assuming that [tex]t_{0}[/tex] is a constant

    [tex]\int_{-\infty}^{+\infty} x(t) \delta (\beta t - t_{0}) dt =
    \int_{-\infty}^{+\infty} e^{at}u(t) \delta (\beta t - t_{0}) dt
    = \int_{0}^{+\infty} e^{at} \delta (\beta t - t_{0}) dt
    = \int_{0}^{+\infty} e^{\frac{a}{b}t_{0}} dt
    = +\infty
  2. jcsd
  3. Jul 11, 2011 #2


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    when you integrate, the delta function is only non-zero at [itex] t_0 [/itex] and
    [tex] \int f(t) \delta (t-t_0) = f(t_0)[/tex]
  4. Jul 11, 2011 #3


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  5. Jul 11, 2011 #4

    hence the result is just

    [tex] e^{\frac{a}{b}t_{0}} [/tex]

    it would be 0 if we didn't integrate from 0 to infinity but from k>0 to infinity


    edit: actually it would not be 0 if the integration was from k>t0/β
    Last edited: Jul 11, 2011
  6. Jul 11, 2011 #5


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    you've got the right idea on the k>t0/b stuff, but i think you need to be careful with the scaling properties of the delta function

    [tex]\int f(t) \delta (\beta t-t_0) dt [/tex]

    now try making the substitution
    [tex] u = \beta t-t_0 [/tex]
  7. Jul 11, 2011 #6
    [tex] u = \beta t - t_{0} => du = \beta dt


    [tex] t = \frac{u + t_{0}}{\beta} [/tex]


    [tex] \int_{-t_{0}}^{+\infty} f(\frac{u+t_{0}}{\beta}) \delta(u) \frac{du}{\beta} [/tex]

    which will give us

    [tex]\frac{1}{\beta} f(\frac{t_{0}}{\beta}) = \frac{e^{\frac{a t_{0}}{\beta}}}{\beta}[/tex] [tex]t_{0}>0[/tex]

    :S, seems complicated.... is this correct? In my previous attempt I forgot to change the variables.. I thought that we just had to search when the function inside [tex]\delta[/tex] is 0, and we just put the [tex]t[/tex] for which δ is 0, in the function before δ..
  8. Jul 11, 2011 #7


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    that looks good to me - you could do it either way if you remember the scaling & translation properties of the delta function.

    Whilst slightly longer I prefer the substitution as you don't need to remember any rules, and we got that extra factor of 1/beta correct without needing to remember it.
  9. Jul 12, 2011 #8
    yes you're right, thanks for your help :)
  10. Jul 12, 2011 #9


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    no worries;)
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