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Integration by change to polar coordinates

  1. Apr 30, 2010 #1
    Hey,

    I have the following integral:
    [tex]

    I = \int_0^\infty dx_1\ldots \int_0^\infty dx_{3N} \int_0^\infty dy_1\ldots \int_0^\infty dy_{3N}\Theta\left(1-\sum_{j=1}^{3N}\left(|x_j|^2+|y_j|^2\right)\right)

    [/tex]
    Now I want to change to polar coordinates by the following substitution:

    [tex]

    \begin{equation}
    x_j &=& r_j \cos \varphi_j;
    y_j &=& r_j \sin\varphi_j
    \end{equation}

    [/tex]
    Then the integral becomes

    [tex]

    \left[\int_0^{2\pi}d\varphi\right]^{3N}\int_0^\infty dr_1\ldots \int_0^\infty dr_{3N}r_1r_2\ldots r_{3N} \Theta\left(1-\sum_{j=1}^{3N}|r_j|^2\right)

    [/tex]
    Now this factors [itex]x_j[/itex] in front of the theta function are bothering me, because if they weren't there, the integral would be simply the volume of sphere with radius 1 in 3N dimensions, and this is a known formula.
    But, they are there so this is a problem...
    Or not, I don't know, is there another maybe easier way to solve this integral?
    Is my substitution perhaps not wel chosen?

    (there are some typo's in the formula, but not important, seems like the preview doesn't renew the latex code)
     
  2. jcsd
  3. Apr 30, 2010 #2

    mathman

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    I don't see any xj only rj, which makes perfect sense.
     
  4. May 1, 2010 #3
    My bad...
    I mean the [itex]r_j[/itex](instead of [itex]x_j[/itex]) coming from the jacobian due to the coordinate transformation.
    I wish they weren't there because the integral would be a volume of a 3N-sphere.

    Is there more advance that can be made otherwise?
     
  5. May 1, 2010 #4

    mathman

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    For better or worse, switching from rectangular to polar coordinates will always look like:

    dxdy = rdrdφ. Intuitively you can look at the r factor being necessary to preserve the dimension of the differential as length squared.
     
  6. May 2, 2010 #5
    Yes, of course.

    Is there maybe anothor way to factorize the original integral in x and y so that I can solve it an a closed form?

    The change to polar coordinates seemed logical to me because I would get a product of two indepent integrals.
     
  7. May 2, 2010 #6

    mathman

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    You hadn't defined Θ, so I couldn't answer.
     
  8. May 3, 2010 #7
    [itex]\Theta[/itex] is Heaviside's stepfunction, equals 1 when it's argument is larger than zero, and 0 otherwise.
    That's why it would be the volume of a sphere with radius 1 when being factorized without the r_j in front
     
  9. May 3, 2010 #8

    mathman

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    Have you tried u_j=r_j2? Then let v=Σu_j and play with the limits.
     
  10. May 4, 2010 #9
    The substitution [itex]r_j^2=u_j[/itex]
    leads,
    for the r-integral only,
    to
    [tex]\frac{1}{2^{3N}}\int_0^\infty du_1\ldots \int_0^\infty du_{3N}\Theta\left(1-\sum_{j=1}^{3N}|u_j|\right)
    [/tex]

    Playing with the limits... I don't see exactly what you mean.
     
  11. May 4, 2010 #10

    mathman

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    I was suggesting a change of variables where w=Σuj replaces one of the u's. Now the problem is adjusting the integration limits to accomodate the change of variable.
     
  12. May 5, 2010 #11
    I still don't get it.
    I think you mean this the following?

    [tex]
    \frac{1}{2^{3N}}\int_0^\infty du_1\ldots \int_0^\infty du_{3N}\Theta\left(1-\sum_{j=1}^{3N}|u_j|\right)

    [/tex]
    with the substitution [itex]w=\sum_j u_j[/itex] you get [itex]dw = \sum du_j[/itex]. How exactly do you enter this differential in the integral then...this become nested integrals then?
     
  13. May 5, 2010 #12

    mathman

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    No! I meant substitute w for (for example) u1, so dw =du1, while all the other u's remain the same. As a result, the problem then becomes setting up the appropriate integration limits.

    One further suggestion (possible alternative) - try setting up the integration limits so that you can dispense with the Heaviside function. Then u1 has limits (0,1), u2 has limits (0,1-u1), u3 has limits (0,1-u1-u2), etc. Then you won't need any change of variables. It looks like a somewhat messy arithmetic problem.
     
  14. May 9, 2010 #13

    mathman

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    I finally found the solution. The surface ∑uk=1 is a simplex in 3N dimensions. Your integral is the volume between 0 and the simplex (with all sides limited to uk ≥ 0). This happens to be a solved problem in simplex analysis (V = 1/(3N)!). It is relatively easy to prove by induction.

    http://en.wikipedia.org/wiki/Simplex
     
  15. May 16, 2010 #14
    Thank you for it!
    I didn't even know it was a known "standard"-integral.
    Even never heard of simplex...
     
  16. May 16, 2010 #15

    mathman

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    A simplex is an n dimensional surface with ∑uk=1 and all uk non-negative.
     
  17. May 19, 2010 #16
    Just curious, but what kind of course is this? I'm finding these advanced calculus topics interesting, but I don't know what they're called. Is it analysis?
     
  18. May 19, 2010 #17

    mathman

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  19. May 19, 2010 #18
    Thanks for the link mathman. What I meant by course was where do people learn about this kind of math? i.e. what's it called? I've taken up to Calc III and diff eq so far, but I haven't seen anything like this.
     
  20. May 20, 2010 #19

    HallsofIvy

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    I've taught linear programming in a PreCalculus course- since it doesn't involve any calculus. However, an "n-Simplex" can be defined as "the convex hull of n+ 1 points" where "convex hull" of any number of points is the set that contains all the points and all straight line segments between them.

    A "1-simplex" is the straight line segment between two points. A "2-simple" is the triangle having three given points as vertices. A "3-simplex" is the tetrahedron having four given points as vertices.
     
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