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Homework Help: 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

    1st. step:
    Consider : [tex]\oint^{\infty}_{-\infty}[/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])

    [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])
     
  2. jcsd
  3. Oct 14, 2008 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Complex Variable---definite integral

    This makes no sense. What contour are you integrating over?

     
  4. Oct 14, 2008 #3
    Re: Complex Variable---definite integral

    um...the pole is at z= iε
    so i find the residule at z=iε
     
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