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Complex Variable-definite integral

  1. Oct 14, 2008 #1
    Complex Variable---definite integral

    Show that
    f(k) = [tex]\frac{1}{2i\pi}[/tex] [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex] =

    1, if k>0
    0, if k<0

    where [tex]\epsilon[/tex] > 0

    The attempt at a solution

    Consider : [tex]\oint[/tex] [tex]\frac{e^{ikz}}{z-i\epsilon}dz[/tex]

    the residule is : e^(-k[tex]\epsilon[/tex])

    so [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex]

    = [tex]{2i\pi}[/tex] e^(-k[tex]\epsilon[/tex])


    f(k) = [tex]\frac{1}{2i\pi}[/tex] [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex] = e^(-k[tex]\epsilon[/tex])


    I can't prove it equal to 0 or 1 and the required condition.
     
  2. jcsd
  3. Oct 14, 2008 #2

    Dick

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    Re: Complex Variable---definite integral

    I hope you mean e^(-k*epsilon). You are leaving out some steps. If you are starting with k>0 you first need to figure out whether you will close the contour in the upper or lower half plane. Since you have exp(ikx) and you want that to be small for x far out on the arc you are using to close the contour, you pick the upper half plane since Im(x)>0 there. Yes, then you pick up the pole at i*epsilon. You are now leaving out the final step of taking epsilon->0. If k<0 then you close in the lower half plane. Now what?
     
  4. Oct 14, 2008 #3
    Re: Complex Variable---definite integral

    So...the above result i obtained means that I just consider k>0, right?

    then I should find the residule for two case:
    k>0 and k<0

    then f(k) is equal to the sum of residules
    correct?
     
  5. Oct 14, 2008 #4

    Dick

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    Re: Complex Variable---definite integral

    You already found the residue of the pole for the k>0 case (though you didn't justify it very well). What does happen if k<0? NO. You don't ADD them. They are two different cases.
     
  6. Oct 14, 2008 #5
    Re: Complex Variable---definite integral

    For k > 0
    [tex]\frac{e^{ikz}}{z-i\epsilon}[/tex]
    pole : z=iε
    residule is e^(-kε)

    so f(k) = e^(-kε)


    for k<0
    [tex]\frac{1}{(e^{imz})(z-i\epsilon)}[/tex]

    where m=|k|

    pole: z=iε
    residule: e^(mε)

    so f(k) = e^(|k|ε)


    umm...how to prove that they equal to 1 or 0???
     
    Last edited: Oct 14, 2008
  7. Oct 14, 2008 #6

    Dick

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    Re: Complex Variable---definite integral

    You might start by listening to what I'm saying. I've said that you need to take the limit of epsilon going to zero. And I've also said you need to think about what your closed contour looks like. What does the closed contour look like in both of these cases? After all, this is contour integration. You'd better start with a contour.
     
  8. Oct 14, 2008 #7
    Re: Complex Variable---definite integral

    contour: consider a semicircle with inf. radius

    that is just wht I have learnt to solve this kind of integral from book
     
    Last edited: Oct 14, 2008
  9. Oct 14, 2008 #8

    Dick

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    Re: Complex Variable---definite integral

    Sort of. That's not very complete or specific. You need a different contour for each of the two cases. What's the difference?
     
  10. Oct 15, 2008 #9
    Re: Complex Variable---definite integral

    now I can prove that the integral is 0 when k< 0

    but....for k>0....i still can;t show that it equals to 1......unless assume that ε → 0....
    but the question just say that ε > 0
     
  11. Oct 15, 2008 #10

    Dick

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    Re: Complex Variable---definite integral

    If you don't take the limit, then the integral is just e^(-k*epsilon). Not 1. But using i*epsilon in a problem like that is almost always a shorthand for 'epsilon>0 and approaching 0'. They may not have bothered to say that explicitly.
     
  12. Oct 15, 2008 #11
    Re: Complex Variable---definite integral

    oic~~~~
    thx~~~
     
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