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Integrating the delta function

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    By using the substitution u=cosx obtain the value of the integral
    [tex]\int[/tex][tex]\delta[/tex](cosx-1/2)dx between 0 and pi


    2. Relevant equations
    I have no idea how to go any further with this apart from substituting in for u!?


    3. The attempt at a solution
     
  2. jcsd
  3. May 19, 2009 #2

    dx

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    What's the integral you get after doing the substitution?
     
  4. May 19, 2009 #3

    D H

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    So, make the substitution! We can't help you if you don't do (or show) your work.
     
  5. May 19, 2009 #4

    tiny-tim

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  6. May 19, 2009 #5
    Sorry, I realise my post was very specific. To be perfectly honest I don't really know where to start. I know that the delta function is defined as
    [tex]\delta[/tex](x-a)=o if x[tex]\neq[/tex]a and tends to infinity is x=a
    shifting property
    [tex]\int[/tex]dx[tex]\delta[/tex](x-a)f(x) = f(a) where the integral is over all space.

    In this case
    I=[tex]\int[/tex][tex]\delta[/tex](cosx-0.5)dx where the integral is between zero and pi
    =[tex]\int[/tex][tex]\delta[/tex](u-0.5)dx letting u=cosx

    not sure how to proceed
     
  7. May 19, 2009 #6

    Cyosis

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    You're substituting, yet after the substitution you integrate over the original variable,dx? Execute the substitution correctly first. Also put the entire expression between [tex] brackets it saves you time and is a lot neater.
     
  8. May 19, 2009 #7

    dx

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    You still have dx in your integral. Write everything in terms of u and du. Don't forget to change the limits of the integral too.
     
  9. May 19, 2009 #8
    Ok...(my brain is obviously not working right now!)
    if u=cosx
    du=-sinxdx
    dx=-du/sinx
    but you need this independent of x? but x=cos^-1(u) ??

    changing the limits:
    when x=0 u=1 and when x=pi u=-1
     
  10. May 19, 2009 #9

    dx

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    You can write sin x = √(1 - u2). So the integral becomes

    [tex] -\int_{1}^{-1} \delta(u - 1/2) \frac{1}{\sqrt{1-u^2}}du [/tex]
     
  11. May 19, 2009 #10

    Cyosis

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    To see that [itex]\sin \arccos u =\sqrt{1-u^2}[/itex] use [itex]\cos^2x+\sin^2x=1[/itex] or use a triangle.
     
    Last edited: May 19, 2009
  12. May 19, 2009 #11

    dx

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    You mean

    [tex] \sin \cos^{-1} u = \sqrt{1-u^2} [/tex].
     
  13. May 19, 2009 #12

    Cyosis

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    Yes I do thanks for pointing it out, it's fixed now.
     
  14. May 19, 2009 #13
    thank you that clarifies it alot!
    Therefore the solution is going to be -1/(1-1/4)^0.5 = 1/(3/4)^0.5 = 2/sqrt(3) ?
     
  15. May 19, 2009 #14

    Cyosis

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    Yep, that's correct.
     
  16. May 19, 2009 #15

    dx

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    Your final answer is correct, but you had a minus sign in your first expression which shouldn't be there.
     
  17. May 19, 2009 #16
    thank you !
     
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