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Straightforward integral?

  1. Jan 25, 2007 #1

    i've seen a textbook where the integration inthe following expression is performed:

    C(S,V,t) = 1-(1/2pi)*exp[-lambda(k0)*t](u(k0,V)/k0^2-i*k0)
    + infin
    *S exp[-.5*kr^2*lambda''(k0)*t]dkr
    - infin

    where kr is k subscript r and k0 is ksubscript 0
    i = complex number
    lamda'' =2nd derivative of lambda



    to me the integral

    + infin
    *S exp[-.5kr^2*lambda''(k0)t]dkr
    - infin

    should give exp[-.5kr^2lambda''(k0)*t]/(-kr*lambda''(k0)*t)

    evaluated at + and - infin and in both cases should go to zero but from the solution above this is clearly incorrect

    please let me know where I am going wrong
  2. jcsd
  3. Jan 26, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\int e^{a k_r}dr= \frac{1}{a}e^{ak_r}[/tex]
    but you have kr2. That square in the exponential means that the anti-derivative cannot be written as an elementary function.
    Are you familiar with
    [tex]\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}[/tex]?
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