Contour integral

  1. [tex]\int \frac{\rho^4 \sin^3{\theta} d \rho d \theta e^{i \rho r \cos{\theta}}}{(2 \pi)^2 [K^2 + \rho^2]}[/tex]

    I am confused about where the singularities are in this function. Will they simply be at \rho = iK and -ik or does the \rho^4 factor make a difference?

    Also the sin^3(\theta) e^(i \rho cos (\theta)) when evaluated on mathematica for example brings in some terms with three different powers of \rho some of which I am afraid will make the function odd in \rho in which case I would struggle with the final contour integration...

    Any help suggestions would be much appreciated.

    Thanks.
     
    Last edited: Feb 19, 2007
  2. jcsd
  3. OK well I just did a test to see if perhaps the factor of \rho^4 makes a difference by dividing the denominator by that factor... it seems that in that case there are no singularities... which can't be right since the integral is supposed to have a finite answer... and a contour integral with no singularities = 0 right? -> confused.

    If I just assume the factor of \rho^4 on the numerator makes no difference, and that the singularities are at -ik and ik, then we still have something that maple says will have terms that aren't even functions in \rho, which I wouldn't know how to deal with when it comes to contour integration.
     
    Last edited: Feb 19, 2007
  4. hmm for now im going to do the theta integral and worry about symmetry and singularities later.
     
  5. Dick

    Dick 25,887
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    Good idea! I think you'll find that the parts that are odd in rho will cancel after the theta integration.
     
  6. We find

    [tex]\int \frac{d \rho}{(2 \pi)^2} \frac{e^{-ir \rho}-e^{ir \rho}}{K^2 +\rho^2} \frac{\rho}{ir} \left [ \rho^2 - \frac{2 \rho}{ir} + \frac{1}{r^2} \right ] [/tex]

    if we leave it like this we find that there are odd terms which I don't know how to handle...

    the only thing I can think of doing is to factor in the rho's into the possibly singular denominator...

    we then find

    [tex]\int \frac{\rho d \rho^2}{(2 \pi)^2} \frac{e^{-ir \rho}-e^{ir \rho}}{ri} \left [ \frac{1}{iK^2 / \rho^2 + 1} + \frac{2}{ir[K^2 / \rho + \rho]} + \frac{1}{r^2[K^2 + \rho^2]} \right ] [/tex]

    Now I'll try and figure out which of these terms exhibit singularities, cancel the ones that don't, and hope I'm doing the right thing...
     
    Last edited: Feb 19, 2007
  7. nrqed

    nrqed 3,083
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    Why are the odd terms a problem? One can still simply do teh contour integration nevertheless, no?
     
  8. nrqed

    nrqed 3,083
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    I will simply use the first form you quoted and write 1/(k^2 + rho)^2 as
    [tex] \frac{1}{(\rho + i K)(\rho -iK)} [/tex] and then you do the integrals over the two exponentials separately, picking up the contour that has a semicircle contribution vanishing at infinity (for example, for [itex] e^{-i r \rho}[/itex] you must close the contour below the real axis in order to pick the pole at -i K).

    Patrick
     
  9. Dick

    Dick 25,887
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    The problem he would have with the odd terms is that the first step to the contour integration is to extend it from the positive real axis to the whole real axis. With the even terms it's easy. Just do it and add a factor of 1/2. What to do with odd (since the extension will necessarily vanish, but not necessarily the original)? Fortunately, if he does the angular integration more carefully (it's not right as stated), he'll find that even power terms in rho are associated with a cos(r*rho) and the odd ones with a sin(r*rho) - so the extension is no problem.
     
    Last edited: Feb 20, 2007
  10. nrqed

    nrqed 3,083
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    Ah ok. The limits of integration were not given and I had assumed that rho was integrated over the real axis but now I realize that those coordinates meant polar coordinates. Thanks for the clarification.
     
  11. :(

    Mathematica gives pretty much the same answer I got (apart from a few factors of 2). There is one odd term with a factor cos(theta), and since the limits of the integration are between theta = 0 to theta = pi, it won't dissapear... I'm really confused and stumped.

    How did you calculate the integral dick? I did it using substitution, the identity cos^2 a + sin^2 a = 1 and then integration by parts twice.

    The answer I'm looking for is

    [tex]\frac{K^2}{2 \pi} \frac{e^{-Kr}}{r} \left [ \frac{3}{Kr} + \frac{3}{K^2 r^2} \right ] [/tex]

    Thanks.
     
    Last edited: Feb 20, 2007
  12. Dick

    Dick 25,887
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    Well, buddy, I just can't agree with you and Mathematica there. The cos(theta) terms should give rise to terms like exp(i*rho*r)+/-exp(-i*rho*r). These are cos(rho*r) and sin(rho*r) respectively, giving me terms containing cos(rho*r)/r^2 and sin(rho*r)/r^3. Both of which are even. So the right answer is out there somewhere! It's not hopeless!
     
  13. This is what I 'm gonna do. I'm gonna go through the integral again, meticulously, on paper. Then transcribe into this thread. Let's see if, and if so where, I went wrong.
     
    Last edited: Feb 20, 2007
  14. Dick

    Dick 25,887
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    Fine idea. If you are REALLY meticulous, you'll probably catch all of your own errors and won't need to transcribe it. That would be pleasant.
     
  15. OK, still pretty much the same answer...

    [tex]G = \int \frac{\rho^4 \sin^3{\theta} d \rho d \theta}{(2 \pi)^2} \frac{e^{i \rho r \cos \theta}}{K^2 + \rho^2}[/tex]

    [tex]= \int D sin^3{\theta} d \rho d \theta e^{a \cos \theta}[/tex]

    where

    [tex]D= \frac{\rho^4}{(2 \pi)^2} \frac{1}{K^2 + \rho^2}[/tex]

    and

    [tex]a=i \rho r[/tex]


    if u = cos(theta), du/dtheta = -sin(theta), so that dtheta=-du/sin(theta),

    [tex]G = -\int D sin^2{\theta} d \rho du e^{a u}[/tex]


    since cos^2(b) + sin^2(b) = 1

    [tex]G = \int D (u^2 - 1) d \rho du e^{a u}[/tex]

    using integration by parts

    [tex]G = \int D \left[ \left ( \frac{u^2 e^{au}}{a} -2 \int \frac{u e^{au}}{a} du \right ) - \frac{e^{au}}{a} \right ] d \rho[/tex]

    [tex]= \int D \left[ \left ( \frac{u^2 e^{au}}{a} -\frac{2}{a} \left( \frac{u e^{au}}{a} - \int \frac{e^{au}}{a} du \right ) \right ) - \frac{e^{au}}{a} \right ] d \rho[/tex]

    [tex]= \int D \left[ \frac{u^2 e^{au}}{a} -\frac{2 u e^{au}}{a^2} + \frac{2 e^{au}}{a^3} - \frac{e^{au}}{a}\right ] d \rho[/tex]

    Considering Dick's next post I'll cut it there for now.
     
    Last edited: Feb 20, 2007
  16. Dick

    Dick 25,887
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    Whoa, stop, stop! You are going backwards!!!! Just evaluate the above expression between u=-1 and u=1. Notice that in this expression the 1/a terms drop out at both limits!!!
     
    Last edited: Feb 20, 2007
  17. Dick

    Dick 25,887
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    Here is your mistake. You've factored the problem as A(u)*B(u). Then to evaluate between u=-1 and u=1, you are computing (A(-1)-A(1))*(B(-1)-B(1)). The correct answer is A(-1)*B(-1)-A(1)*B(1). Which is NOT THE SAME THING!!!!
     
  18. Yes I was making a mistake putting in the limits. Thanks.

    OK so now we have

    [tex]= \int D \left[ \frac{2 e^{-a}}{a^2} +\frac{2 e^{-a}}{a^3} + \frac{2 e^{a}}{a^2} - \frac{2 e^{a}}{a^3}\right ] d \rho[/tex]

    but since D is even in rho and the exponential is odd in rho, the terms with a factor of 1/a^2 (which is even in rho) will be odd in rho, (with the terms having a factor of 1/a^3 being even in rho).
     
  19. Dick

    Dick 25,887
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    Beg to differ. The a^2 term looks like (exp(a)+exp(-a))/a^2. I call that EVEN in rho. And (exp(a)-exp(-a))/a^3 is also even. Flip a to -a and see what happens... Exponentials are NOT ODD.
     
    Last edited: Feb 20, 2007
  20. nevermind........
     
    Last edited: Feb 20, 2007
  21. woo!!! I think i just understood what you meant. I was just too dense to see. if we convert the exponentials into sin and cos we are sorted! thanks a million dick, should be fine now :)
     
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