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Time-shifted Integral with Unit Impulse Function

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral.
    [tex] \int_{-\infty}^\infty f(2 -t) \delta(3 -t) \, dt [/tex]

    2. Relevant equations
    The Unit Impulse function is defined
    [tex] \hspace{18mm} \delta(0) = 0 \hspace{10mm} t \neq 0 [/tex]
    [tex] \hspace{-15mm} \int_{-\infty}^\infty \delta(t) \, dt = 1 [/tex]

    3. The attempt at a solution
    Using a u-substitution ##u = 3 -t##, ##dt = -du##, ##t = 3 -u##
    [tex] \int_{-\infty}^\infty f(2 -t) \delta(3 -t) \, dt [/tex]
    [tex] = \int_{-\infty}^\infty f(2 -(3 -u)) \delta(u) (-1) \, du [/tex]
    [tex] = -\int_{-\infty}^\infty f(u -1) \delta(u) \, du [/tex]
    [tex] = -f(0 -1) [/tex]
    [tex] = -f(-1) [/tex]

    Please let me know if the above is unclear. My question is 1) Is this the right solution? and 2) Trusted sources tell me that the answer is just ##f(-1)##. Which is the correct answer? Is there some mathematical detail I am missing? Hope this is clear, and hope you can help. ##\mathcal{Thanks}##
     
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  3. Sep 14, 2015 #2

    DEvens

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    You did a substitution of u = 3-t. And so you got a du instead of a dt. Good. What happens to the limits on the integral?
     
  4. Sep 14, 2015 #3

    Orodruin

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    Yes, you have changed the integration limits when you did the variable substitution. ##u\to -\infty## when ##t\to +\infty##
     
  5. Sep 14, 2015 #4
    For the lower limit of the integral, since ##t \rightarrow -\infty##, ##u \rightarrow 3 -(-\infty) \rightarrow 3 +\infty \rightarrow \infty##, and in the upper limit of the integral, ##t \rightarrow \infty##, implying ##u \rightarrow 3 -\infty \rightarrow -\infty##. The net effect is to annihilate the ##-1## in front of the integral, giving the evaluation ##f(-1)##. Thank you both for the insight!
     
  6. Sep 14, 2015 #5

    DEvens

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    Alternatively, you could use the fact that the Dirac delta is even. That means ##\delta(x) = \delta(-x)##. So you could just get rid of the annoying minus sign.

    https://en.wikipedia.org/wiki/Dirac_delta_function
     
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