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Delta Function Integration

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data


    [tex]\int_{-3}^{5} e^{-2t} sin(t-3) \delta(t-5) dt[/tex]

    2. Relevant equations

    [tex]\int_{-\infty}^{\infty} f(t) \delta(at-t_0) dt = \frac{1}{|a|}f(\frac{t_0}{a})[/tex]

    3. The attempt at a solution

    [tex]e^{-2(5)} sin (5-3) = e^{-10} sin (2)[/tex]

    The solution given by the professor was:

    [tex]\frac{1}{2} e^{-10} sin (2)[/tex]

    I don't understand where he got the [tex]\frac{1}{2}[/tex] from.

    If anyone could help me it would be greatly appreciated.
  2. jcsd
  3. Sep 26, 2007 #2
    It's a bit cheeky, and technically your professor's answer is contestable. Because your integration limits are only up to 5, and the delta function is delta(t-5), you've only integrated "half" of the delta spike. Now, if that sounds absurd -- "how the hell can you only integrate half of an infinitesimally thin spike?!" -- just remember that the delta function itself is not really very sensible as a function anyway. Personally, I treat that sort of integral as badly defined; more specifically, since I'm a physicist, I arrange my physics so that the maths never requires me to contemplate these soul-searching issues :wink:
  4. Sep 26, 2007 #3
    Hm, interesting... I never knew that you integrate only half the delta spike.
  5. Sep 26, 2007 #4


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    You shouldn't. As genneth was pointing out, there are perfectly fine representations of the delta function in which you can get a completely different answer. The correct answer is 'undefined', with all due respect to the composer of the solutions.
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