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Difficult integral

  1. Feb 22, 2008 #1
    I fear that this one is really hard, if not impossible, but an analytic answer would be way more usefull than a numerical one. Who can help me in the right direction?




    [itex]\int_0^{arccos(a)} d\phi \frac{cos(\phi)}{(cos(\phi)+\sqrt{cos^2(\phi) - a^2})(x \sqrt{cos^2(\phi)-a^2}+y sin(\phi) + z + x_0 cos(\phi) )}[/itex]

    with 0<a<1 and the phi integral only over positive values of the squareroot

    Approximations for y=0 and a small are also welcome.

    This integral comes from the double integral [itex]\int_0^{\infty} dk\int_0^{2\pi}d\phi \frac{cos(\phi)}{(cos(\phi)+\sqrt{cos^2(\phi) - a^2}} e^{i k(x \sqrt{cos^2(\phi)-a^2}+y sin(\phi) + z + x_0 cos(\phi) )}[/itex]
     
    Last edited: Feb 23, 2008
  2. jcsd
  3. Feb 22, 2008 #2
    i dont understan anything of what you wrote.
    could you post somthing more readible...

    use the buttons if you dont understand tex code.

    ciao
    marco
     
  4. Feb 22, 2008 #3
    Yeah i agree with u! i am staring at it for a few minutes but it is quite hard to dechiper, quite ambiguous!
     
  5. Feb 22, 2008 #4

    Hurkyl

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    Ambiguous? I don't think that word means what you think it means. :tongue: The expression is written in the syntax of Mathematica. (and I'm pretty sure it's syntactically correct)
     
  6. Feb 22, 2008 #5
    Yeah i guess! I just wanted to say that it is hard to read what the OP posted, but i guess i said the wrong way!
     
  7. Feb 25, 2008 #6
    Posted something more readable. Wasn't aware of the Tex possibilities at first, so I used the Mathematica syntax.
     
  8. Feb 25, 2008 #7
    Let's hope someone will take the time to work this out, cuz it really looks nasty!!!
     
  9. Jan 8, 2009 #8
    Probably, unfamiliarity with the advanced topic is creating the problem. But, as I know, this particular integral is of Fourier Integral Theorem, and it solved from the left to the right i.e. formal tactics of a regular double integration not necessarily followed. This means, this integral does not generates a surface of the solid, but rather simplifies the wave equation from amplitude to frequency domain. Or the same analogy holds also for any sinusoidal signal wave. The original question was not that complex, when all variables get converted into spherical or cylindrical parametric form, it looks quite messy, but in reality, it is only a tedious job, not genuine.
     
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