1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

  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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    no worries;)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook