Complicated Double Integral

  • Thread starter Hendrik
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  • #1
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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:

Answers and Replies

  • #2
6
0
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
 
  • #3
6
0
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
6
0
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
benorin
Homework Helper
Insights Author
Gold Member
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80
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.
 
  • #6
6
0
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
 
  • #7
HallsofIvy
Science Advisor
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956
You set n= 1, not 0.
 
  • #8
Hurkyl
Staff Emeritus
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Have you looked at it in rectangular coordinates? It looks as if it would be much easier.
 
  • #9
benorin
Homework Helper
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Gold Member
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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?
 
  • #10
6
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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
 

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