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I Prove that ∫f(x)δ(x)dx=f(0)

  1. Jan 22, 2017 #1
    I somehow understand "why" that works, but when it comes to prove it I fail... I managed to "half-prove" it by
    saying:
    ∫f(x)δ(x)dx=f(a)*δ(a)+...+f(0)*δ(0)+...+f(b)*δ(b)=f(a)*0+...f(0)*1+...+f(b)*0=f(0)
    note: integral goes from "a" to b", a<0, b>0 and δ(x) is Dirac-delta function
    but I figured out I've missed to put Δx in every term and now it confuses me even more...
    help!
     
  2. jcsd
  3. Jan 22, 2017 #2

    PeroK

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    What's your definition of ##\delta##?
     
  4. Jan 22, 2017 #3
    it equals 1 when x=0, and it equals zero when x is different than zero (Dirac function)
     
  5. Jan 22, 2017 #4

    PeroK

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    That's not the Dirac function. That's a function that is equivalent to the zero function when integrating over an interval.
     
  6. Jan 22, 2017 #5
    ahhhhh yeah you are right... at zero dirac function equals +inf and integral from -inf to +inf of Dirac function equals one by definition. how I'm gonna attack this problem now or better to say how I'm gonna attack it anyway?
     
  7. Jan 22, 2017 #6

    PeroK

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    It really depends on your starting point in defining the Dirac delta function.
     
  8. Jan 22, 2017 #7
    if I say I want to prove it in a manner similar from the first post (by using definition of definite integral) then what "type of definition" of Dirac function should I use?
     
  9. Jan 22, 2017 #8

    PeroK

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    You could assume that it's a function whose integral about the point 0 is 1, no matter how small the interval containing 0.

    You could assume ##f## ia continuous at 0 as well.
     
  10. Jan 22, 2017 #9
    well, now I think my high-school math skills aren't sufficient to actually prove it.... let me ask if you could prove it in this definite integral fashion:
    ∫f(x)δ(x)dx=f(a)*δ(a)+...+f(0)*δ(0)+...+f(b)*δ(b)=f(a)*0+...f(0)*1+...+f(b)*0=f(0) cause that's the furtherst I "understand". thanks!
     
  11. Jan 22, 2017 #10

    PeroK

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    If you're at high school level I wouldn't worry about this sort of thing. Instead take this result as essentially a postulated property of the delta function: that when integrated with another function it picks out the function value at a specific point.
     
  12. Jan 22, 2017 #11
    actually I'm at college where math is frozen to math from high school (don't ask me why, I hate it). no matter what, as always, I wanted to go step further in my understanding. we have a DSP (Digital Signal Processing) subject and actually I have to self-study it... so tell me what kind of math I need in order to decompose this problem and actually "understand" it. if it is a calculus based proof (at any levels) you can post it or give me names/external links to any other non-calculus based proofs. I wanna give it a try. and yes, I will be very thankful!
     
  13. Jan 22, 2017 #12

    PeroK

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    The Dirac delta function is an exception in that it's not a function and it is difficult to use rigorously. Neither straight calculus nor real analysis will help. It's use in physics as an infinite spike function is invaluable and I would accept it as a unique creature for this reason.

    I think you'd gain relatively little from the time and effort needed to study it formally. Time that would be better spent on calculus, linear algebra or differential equations.
     
  14. Jan 22, 2017 #13
    thank you for your guidance! I always wondered why they put any of these subjects before you gain deep understanding of every math concept behind it. I hope that's not the case with universities around the world.
     
  15. Jan 22, 2017 #14

    micromass

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    The property in the OP is actually the definition of the Dirac delta function.
     
  16. Jan 22, 2017 #15

    FactChecker

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    Since the delta function is 0 except at x=0, you can simplify the integrand to get rid of all f(x) except f(0). Work from there.
     
  17. Jan 22, 2017 #16

    micromass

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    I don't get why everybody is trying to prove this. It cannot be proven. It's a definition.
     
  18. Jan 22, 2017 #17

    FactChecker

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    His definition has no function f(x). That is very common.
     
  19. Jan 22, 2017 #18

    micromass

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    Common but wrong. The property cannot be proven from his definition.
     
  20. Jan 22, 2017 #19

    FactChecker

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    I could prove it as long as integration retains its linear properties.
     
  21. Jan 22, 2017 #20

    micromass

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    Then I'm sorry to say that your proof is wrong.
     
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