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Integral of exponential absolute functions

  1. Sep 20, 2010 #1

    I am having difficulty solving the following integral:


    I have tried to use an explicit form of the absolute, eg.

    -(a|x|+ikx) = \left\{\stackrel{-(ik+a)x\ x>0} {-(ik-a)\ x<0}

    Does this allow me to seperate the integral into a sum of two integrals?


    This was my best guess, but the result I got did not converge, so either I did the integral improperly, or else this is not a legal method.

    Would someone be so kind as to share their knowledge?
    Last edited: Sep 20, 2010
  2. jcsd
  3. Sep 20, 2010 #2


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    Yes, you can but I think a better reduction is
    [tex]e^{-a|x|+ ikx}dx= e^{-a|x|} e^{ikx}= e^{-a|x|}(cos(kx)+ i sin(kx))[/tex]
    Then make use of the symmetries involved.
    (I am assuming you mean that to be an exponent. You missed the "^" in your LaTex.)
  4. Sep 20, 2010 #3
    Thank you.

    So, by symmetry, the sin term dissappears, and the rest of the integral can be taken as twice the integral from zero to infinity?
  5. Sep 20, 2010 #4


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    Yes. In your original separation into two integrals, the exponent in the first term is wrong. It should be (a-ik)x. You had it right on the previous line.
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