1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complicated Double Integral

  1. Feb 18, 2006 #1
    Hi,
    I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.

    Consider

    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} { \sqrt{ 4z^2+4r^2+4r\cos(\theta)+2-4z }^(2n+3) }\,dr\,d\theta [/tex]

    which is a function of z for given n, n>0.
    The problem is that I need an analytical sloution, because \psi_n shall be integrated again which can then be done numerically. I considered basic integration methods and gave the expression to maple but it didn't help. I wonder if there is any possibility to simplify the integrand / solve the integral.

    If you don't think so please tell me so, too, this would already be some help. Thank you.

    Hendrik
     
    Last edited: Feb 18, 2006
  2. jcsd
  3. Feb 18, 2006 #2
    Hi,
    I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.

    Consider

    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} {\sqrt{4z^2+4r^2+4r\cos(\theta)+2-4z}^{2n+3}}\,dr\,d\theta [/tex]

    which is a function of z for given n, n>0. The problem is that I need an analytical sloution, because \psi_n shall be integrated again which can then be done numerically. I considered basic integration methods and gave the expression to maple but it didn't help. I wonder if there is any possibility to simplify the integrand / solve the integral.

    If you don't think so please tell me so, too, this would already be some help. Thank you.

    Hendrik
     
  4. Feb 18, 2006 #3
    Hi,
    I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.

    Consider

    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} {sqrt{4z^2+4r^2+4r\cos(\theta)+2-4z}^{2n+3}}\,dr\,d\theta [/tex]

    which is a function of z for given n, n>0. The problem is that I need an analytical sloution, because \psi_n shall be integrated again which can then be done numerically. I considered basic integration methods and gave the expression to maple but it didn't help. I wonder if there is any possibility to simplify the integrand / solve the integral.

    If you don't think so please tell me so, too, this would already be some help. Thank you.

    Hendrik
     
  5. Feb 18, 2006 #4
    Hi,
    I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.

    Consider

    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} {\sqrt{4z^2+4r^2+4r\cos(\theta)+2-4z}^{2n+3}}\,dr\,d\theta [/tex]

    which is a function of z for given n, n>0. The problem is that I need an analytical sloution, because \psi_n shall be integrated again which can then be done numerically. I considered basic integration methods and gave the expression to maple but it didn't help. I wonder if there is any possibility to simplify the integrand / solve the integral.

    If you don't think so please tell me so, too, this would already be some help. Thank you.

    Hendrik
     
  6. Feb 19, 2006 #5

    benorin

    User Avatar
    Homework Helper

    Did you mean this ?

    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} {\sqrt{4z^2+4r^2+4r\cos (\theta)+2-4z^{2n+3}}}\,dr\,d\theta [/tex]


    If so...


    [tex] \psi_n(z) =(z-\frac{1}{2}) \int_0^{2\pi}\int_0^1 \frac{ (r \cos(\theta) + \frac{1}{2})^n \cdot r} {\sqrt{4z^2+4r^2+4r\cos (\theta)+2-4z^{2n+3}}}\,dr\,d\theta [/tex]

    for a start, try interchanging the order of integration the stuff in z (under the radical) is just a constant so collect it as one, expand the numerator as a binomial series and integrate termwise after completing the square in the denominator. (maybe that'll work: try it.
     
  7. Feb 20, 2006 #6
    Thanks for the answer, but no, I meant:

    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} {\sqrt{4z^2+4r^2+4r\cos(\theta)+2-4z}^{2n+3}}\,dr\,d\theta [/tex]

    ...sorry for the latex trouble. Taking out the constants is a good idea and it might speed up the numerical processing a little. But the question remains if the integral is analytically treatable, even for n=1.

    Let's talk about this guy here
    [tex] \psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0)^n \cdot r} {\sqrt{r^2+r\cos(\theta)+p}^{2n+3}}\,dr\,d\theta [/tex]
    and set n=0. We obtain:
    [tex] \psi_1(z) = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0) \cdot r} {\sqrt{r^2+r\cos(\theta)+p}^{5}}\,dr\,d\theta [/tex]

    What do you think?
    Hendrik
     
  8. Feb 20, 2006 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You set n= 1, not 0.
     
  9. Feb 20, 2006 #8

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Have you looked at it in rectangular coordinates? It looks as if it would be much easier.
     
  10. Feb 21, 2006 #9

    benorin

    User Avatar
    Homework Helper

    Does

    [tex] \psi_1(z) = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0) \cdot r} {\sqrt{r^2+r\cos(\theta)+p}^{5}}\,dr\,d\theta = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0) \cdot r} {\left( \sqrt{r^2+r\cos(\theta)+p}\right) ^{5}}\,dr\,d\theta = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0) \cdot r} {\left( r^2+r\cos(\theta)+p}\right) ^{\frac{5}{2}}}\,dr\,d\theta[/tex]

    or what?
     
  11. Mar 7, 2006 #10
    Hi Benorin, Hi Hurkyl,

    meanwhile I solved the problem using mathematica instead of maple which I found out is much more performant numerically. I still don't know if it works other way, but thank you guys, anyway.

    Hendrik
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Complicated Double Integral
Loading...