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Evaluating integral with delta function (Fourier Transform)

  1. Nov 4, 2015 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    Evaluate the Following integrals

    1. http://www4b.wolframalpha.com/Calculate/MSP/MSP10141fif9b428c5bab0b00005dc489hi851d28h7?MSPStoreType=image/gif&s=37&w=164.&h=35. [Broken]


    2. Relevant equations
    http://www4b.wolframalpha.com/Calculate/MSP/MSP23802098254f7bhefdhd00002e1e8be006773e70?MSPStoreType=image/gif&s=64&w=184.&h=35. [Broken]

    3. The attempt at a solution

    I don't know. please help. I don't know if the -3 is correct. it is what wolfram gave me when I typed it in to paste it here. If it is correct I don't know why. I would think the x being cubed would affect the answer so that it isn't just -a
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 4, 2015 #2

    grandpa2390

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    and http://www4f.wolframalpha.com/Calculate/MSP/MSP18161hebeg6i2g9e2cbi0000626gc3eg86i0g2ab?MSPStoreType=image/gif&s=16&w=290.&h=35. [Broken] but replace the y's with x'
    once again I don't know if the result given by wolfrahm is correct. I don't understand how to break this down.
     
    Last edited by a moderator: May 7, 2017
  4. Nov 4, 2015 #3

    PeroK

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    All you have to do is to work out when the argument for the ##\delta## function is ##0##. Can you do that for the two examples?
     
  5. Nov 4, 2015 #4

    grandpa2390

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    well that would be when x=0 right? δ(0)=0

    so I would plug that into the f(x)? and get f(0^3-3)= -3
    and -3^3= -27

    for the next one

    δ(y-x) = 0 when y=x
    so: (x-3)^3

    that's pretty simple. thanks!
     
  6. Nov 4, 2015 #5

    Ray Vickson

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    Why would you think that? ##\int_{-\infty}^{\infty} f(x) \delta(x-a) \, dx = f(a)## [Note: NOT what you wrote under Heading 2.!] So, for ##a = 0## and ##f(x) = x^3 -3## we get the answer ##f(0) = (0^3 - 3)##.
     
    Last edited by a moderator: May 7, 2017
  7. Nov 4, 2015 #6

    grandpa2390

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    They are both valid. My professor gave us both. to make my question less confusing and wordy, I wrote the one that I thought applied. but they are both different ways of expressing the same thing.

    find where δ(x) = δ(0)

    at x=0
    f(0-a) = f(-a)

    or in the equation you posted, x=a and f(x)=f(a)

    it's the same thing. apply it to my problem. the same way.

    so we get 0^3-3=-3
     
    Last edited: Nov 4, 2015
  8. Nov 4, 2015 #7

    Ray Vickson

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    Sorry: I mis-typed it. What I meant was ##\int_{-\infty}^{\infty} f(x) \delta (x-a) \, dx = f(a)##, not what I wrote. Your equation was correct, and I had made a stupid mistake.
     
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