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Delta function integral

  1. Jan 17, 2007 #1
    let [tex]f(y)=\int_0^2 \delta(y-x(2-x))dx[/tex]. Find f(y) and plot it from -2 to 2.

    I know how to calculate [tex]\delta (g(x))[/tex] but i am not sure how to treat it with the y. I thought possibly to solve the quadratic in the delta function to find what x will equal for the roots in terms of y and got 1+(1-y)^1/2 and 1-(1-y)^1/2. i am not sure though how to find f(y) with the integral. any suggestions?
     
  2. jcsd
  3. Jan 18, 2007 #2

    arildno

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    Treat the cases separately:
    1. The singularity lies in the open interval between the end points
    2. The singularity lies outside the interval
    3. The singularity lies at an end point
     
  4. Jan 18, 2007 #3

    HallsofIvy

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    The singularity arildno refers to is where y- x(2-x)= 0. In other words when y= 2x- x2. Find the value of that integral, f(y), for y in the positions arildno listed and that's your function.
     
  5. Jan 18, 2007 #4
    so what youre saying is to form the integral [tex]f(y)=\int 2x-x^2 dx[/tex]?
    if its outside the interval do i let the limits be (-infinity to 0) and (2 to infinity)
    for inside 0 to 2 but at the endpoints what are the limits?
     
  6. Jan 19, 2007 #5

    arildno

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    No, why do you think that? :confused:
     
  7. Jan 19, 2007 #6

    HallsofIvy

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    What is [itex]\int_a^b \delta(x)dx[/itex] for any a, b?
     
  8. Jan 19, 2007 #7

    D H

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    Even more appropriately,

    What is [itex]\int_a^b \delta(x)f(x)dx[/itex] for any a, b?

    One way to solve the problem at hand is to convert the original problem,
    [tex]f(y)=\int_0^2 \delta(y-x(2-x))dx[/tex]
    to the form
    [tex]f(y)=\int_{u_0}^{u_1} \delta(u)f(u)du[/tex]

    Note that the integrand may have two singularities between the integration limits, depending on the value of y.
     
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