# Complicated Double Integral

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

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

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:

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

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

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

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

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

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

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

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

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

benorin
Homework Helper
Gold Member
Did you mean this ?

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

If so...

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

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.

Thanks for the answer, but no, I meant:

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

...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.

$$\psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0)^n \cdot r} {\sqrt{r^2+r\cos(\theta)+p}^{2n+3}}\,dr\,d\theta$$
and set n=0. We obtain:
$$\psi_1(z) = \int_0^{2\pi}\int_0^1 \frac{(r \cos(\theta)+x_0) \cdot r} {\sqrt{r^2+r\cos(\theta)+p}^{5}}\,dr\,d\theta$$

What do you think?
Hendrik

HallsofIvy
Homework Helper
You set n= 1, not 0.

Hurkyl
Staff Emeritus
Gold Member
Have you looked at it in rectangular coordinates? It looks as if it would be much easier.

benorin
Homework Helper
Gold Member
Does

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

or what?

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