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Simple integral in Mathematica

  1. Aug 23, 2011 #1
    The integral is :

    [PLAIN]http://i038.radikal.ru/1108/ed/a6cd0c5e8f36.jpg [Broken]

    , where "y" - coordinate, "kx" - component of wave vector

    if we calculate this integral on a paper with a pen and using compex variables theory we obtain:

    1) y>0

    here, in the capacity of contour we take semicircumference in the upper half plane

    [URL]http://s006.radikal.ru/i215/1108/c9/835aa4a65f60.jpg[/URL]

    2) y<0

    here, in the capacity of contour we take semicircumference in the lower half plane

    [URL]http://s58.radikal.ru/i162/1108/b7/37bb3bc6170d.jpg[/URL]

    So the total answer is

    [URL]http://s004.radikal.ru/i208/1108/4f/c35aafd1c20b.jpg[/URL]

    Let's calculate it in Math now.

    [URL]http://s40.radikal.ru/i090/1108/f8/cf9f67043005.jpg[/URL]

    I wonder where Math gets "-" in Exp[] ?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 23, 2011 #2

    Dale

    Staff: Mentor

    The - sign is correct. You made an error doing it by hand. To check do a numeric integration and compare to the analytical answers.
     
  4. Aug 23, 2011 #3
    ok, but where is an error?
     
  5. Aug 24, 2011 #4
    Looks to me you got the y's mixed up. When the semicircle is over the upper half-plane, I obtain an integral asymptotic to:

    [tex]\int_{0}^{\pi} e^{R\sin(t) y} dt[/tex]

    and that converges when y<0. Similar dif for the lower half-plane.
     
  6. Aug 24, 2011 #5
    i'm sorry, but where is "-" in Exp[] ?
     
  7. Aug 24, 2011 #6
    Let [itex]q=Re^{it}[/itex] over the semi-circle in the upper half-plane, and I want to know the bounds on the integral. So, I can write that it's less than:

    [tex]\int_0^{\pi}\left|\frac{e^{-iRe^{it}y}}{R^2e^{2it}+k^2}Rie^{it}\right|dt[/tex]

    and really, it's going to be dominated by the real part of the exponent in the numerator right? So asymptotically, I (think), it's going to approach:

    [tex]\int_0^{\pi}\frac{e^{-iRe^{it}y}}{R}dt[/tex]

    [tex]\int_0^{\pi}\frac{e^{-iR(\cos(t)+i\sin(t))}}{R}dt[/tex]

    So now, I'm only interested in the absolute value of that and that's dependent on it's real part:

    [tex]\int_0^{\pi}e^{R\sin(t)}dt[/tex]

    But I did that really quick and sloppy. Would need to double-check it and do a better job with inequalities and all if I were turning it in for a grade.
     
  8. Aug 24, 2011 #7
    you are right, i'm sorry. thanks!
     
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