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Is this integration correct?

  1. May 21, 2015 #1
    I have come across the following integral in my PhD research.

    [tex]\int_{-L_y/2}^{L_y/2}\int_{-L_x/2}^{L_x/2}e^{2 i k \sqrt{(d \cos\theta + x)^2+(d \sin\theta + y)^2}}dxdy[/tex]

    Ultimately, x and y are integrated out and I will be left with a function of ##\theta##. I am doing this the following way. First I make the following substitutions.

    ##u=d\cos\theta+x \rightarrow du=dx ##
    ## v=d\sin\theta+y \rightarrow dv=dy ##

    So then I get,
    [tex]\int \int e^{2 i k \sqrt{u^2+v^2}}dudv.[/tex]

    But then I can make yet another variable substitution by letting,

    ## r= \sqrt{u^2+v^2} \rightarrow dudv=rdr. ##

    Which gives me,

    [tex]\int e^{2 i k r}r dr[/tex]

    I can easily integrate this by parts, and (assuming I did all my steps correctly) I get the following.

    [tex]\frac{1}{2k}e^{2 i k r}\left[ \frac{1}{2k}-r \right][/tex]

    Now here is where it gets tricky for me

    I now need to back substitute for ##r## to get it in terms of my original variables ##x## and ##y##. When it comes to evaluating the limits, can I just use the expression I got (after substituting back) and first evaluate for ##x## and then for ##y##?

    Or perhaps I can just leave r in and just use the min and max values of u and v to write the limits of r?

    ##r_{min}=\sqrt{(d\cos\theta - L_x/2)^2+(d\sin\theta-L_y/2)^2}##
    ##r_{max}=\sqrt{(d\cos\theta + L_x/2)^2+(d\sin\theta+L_y/2)^2}##
     
  2. jcsd
  3. May 21, 2015 #2

    RUber

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    Your method looks good. I didn't check the integration by parts, but I don't doubt that it is correct.
    However, when you go to put the limits of integration, it is best to consider the limits with each change of variable.
    ## u = d\cos\theta + x## so your limits on ##u## are simply ## d\cos\theta-L_x/2 , d\cos\theta+L_x/2 ## similarly for ##v##.
    When you change to ##r##, you have to look at the actual minimum value of the variable over the range of u and v.
    If ##L_x/2 > d\cos \theta \text{ and } L_y/2 > d\sin \theta## then your minimum will be 0. Consider various other cases.
    Your maximum will be at ##d \cos \theta + (sign [d\cos\theta]) L_x/2## and similarly for the y component.
    Know the signs of your constants and the domain of theta to make this process simpler.
     
  4. May 21, 2015 #3

    RUber

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    ## \int e^{2ikr} r dr##
    ## \frac{1}{2k}e^{2 i k r}\left[ \frac{1}{2k}-r \right]##


    When I checked the integration by parts, I still had some imaginary coefficients, like ##\frac{1}{2ik}##.
     
  5. May 21, 2015 #4
    Thank you for your response! That makes perfect sense. I will go through it and see if it is straight forward or not.

    Also yes you are correct, I accidentally dropped a factor of ##i##.
     
  6. May 25, 2015 #5
    $$ dx\,dy = r\,dr\,d\phi $$
     
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