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Math Help - Dirac Delta Function Integral

  1. #1
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    Dirac Delta Function Integral

    Hi.

    Im having some trouble with the following question:

    "Evaluate the integral: e^x [dirac(x + 1) − dirac(x − 2)] dx from -ve infinity to 1"

    Am i supposed to use a Laplace transformation? In which case I split it into 2 integrals:

    e^x * dirac(x + 1)

    minus....
    e^x * dirac(x − 2) dx

    So for the first integral I get exp(p-1), and for the second exp(2-2p). So overall exp(p-1) - exp(2-2p).

    However as the original integral was from -ve infinity to 1, the second dirac integral (at x=2) would be out of this boundary. So would I disregard this?


    Have I gone about this problem in the correct way?

    Thanks in advance.
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  2. #2
    MHF Contributor

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    Why in the world would you use a "Laplace Transform"? \int_{-\infty}^1 e^x[\delta(x+ 1)- \delta(x- 2)]dx= \int_{-\infty}^1 e^x\delta(x+1) dx- \int_{-\infty}^1 e^x\delta(x- 2)dx

    The definition of the Dirac delta function is that it is the functional that assigns to every function f(x) the value \int_a^b f(x)\delta(x)dx= f(0) if 0 is in the interval [a, b], 0 otherwise. A simple change of variable extends that to \int_a^b f(x)\delta(x- q)dx= f(q) if q is in the interval [a, b], 0 otherwise.

    The first integral involves \delta(x+ 1)= \delta(x-(-1)) and -1 is in the interval (-\infty, 1]. The integral is e^{-1}.

    The second integral involves \delta(x- 2) but 2 is NOT in the interval (-\infty, 1]. The integral is 0.

    I have no idea where your "p" came from.
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