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jjbyram
 Post subject: very difficult definite integral
PostPosted: Mon, 9 Apr 2012 16:31:57 UTC 
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The following integral can be used to calculate the moment of intertia of the nucleus of an atom. How do I go about solving the below Moment of inertia integral?
(It is very complicated because of a variable density)

I= ρ* ∫∫∫ r^4*sin^3(θ)* [1+e^((r-R)/a)]^(-1) dr dθ dφ

dr is integrated from 0 to R

dθ is integrated from 0 to pi

dφ is integrated from 0 to 2pi

Any advice, or hints would be greatly appreciated. Are there number calculators that could solve this?
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outermeasure
 Post subject: Re: very difficult definite integral
PostPosted: Mon, 9 Apr 2012 16:51:43 UTC 
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jjbyram wrote:
The following integral can be used to calculate the moment of intertia of the nucleus of an atom. How do I go about solving the below Moment of inertia integral?
(It is very complicated because of a variable density)

I= ρ* ∫∫∫ r^4*sin^3(θ)* [1+e^((r-R)/a)]^(-1) dr dθ dφ

dr is integrated from 0 to R

dθ is integrated from 0 to pi

dφ is integrated from 0 to 2pi

Any advice, or hints would be greatly appreciated. Are there number calculators that could solve this?


The \theta- and \varphi-integrals are easy. The r-integral is not expressible in closed form using elementary functions alone (you can play with polylog if you want...).

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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jjbyram
 Post subject: Re: very difficult definite integral
PostPosted: Mon, 16 Apr 2012 19:05:41 UTC 
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Supposed the integral was simplified:

I= ρ* ∫∫∫ r^4*sin^3(θ)* [1+e^(r/a)]^(-1) dr dθ dφ

dr is integrated from 0 to R

dθ is integrated from 0 to pi

dφ is integrated from 0 to 2pi

Can this now be solved with integration by parts?

I saw that on a lookup table:

∫ [a+be^(px)]^(-1) dx = x/a - (1/ap)*log(a+be^px)

if so, how would the integration of parts be set up?

Most appreciatively,

Jeff
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jjbyram
 Post subject: A nasty u-substitution integral?
PostPosted: Tue, 17 Apr 2012 20:18:00 UTC 
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I am tryining to solve the following integral for I, the moment of inertia ( ρ and a are constants):

I= ρ* ∫∫∫ r^4*sin^3(θ)* [1+e^(r/a)]^(-1) dr dθ dφ

dr is integrated from 0 to R

dθ is integrated from 0 to pi

dφ is integrated from 0 to 2pi

Can this now be solved with integration by parts?

I saw that on a lookup table:

∫ [ [a+be^(px)]^(-1) ] dx = x/a - (1/ap)*log(a+be^px)

if so, how would the integration of parts be set up?

Most appreciatively,

Jeff

:confused:
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Shadow
 Post subject: Re: A nasty u-substitution integral?
PostPosted: Tue, 17 Apr 2012 21:50:02 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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jjbyram wrote:
I am tryining to solve the following integral for I, the moment of inertia ( ρ and a are constants):

I= ρ* ∫∫∫ r^4*sin^3(θ)* [1+e^(r/a)]^(-1) dr dθ dφ

dr is integrated from 0 to R

dθ is integrated from 0 to pi

dφ is integrated from 0 to 2pi

Can this now be solved with integration by parts?

I saw that on a lookup table:

∫ [ [a+be^(px)]^(-1) ] dx = x/a - (1/ap)*log(a+be^px)

if so, how would the integration of parts be set up?

Most appreciatively,

Jeff

:confused:


First use the constant limits to switch the order and do the \theta integral first, replace a \sin^2\theta with 1-\cos^2\theta and then let u=-\cos\theta so that du=\sin\theta\,d\theta, then do the \varphi integral next since there's no \varphi dependence, that comes out immediately. For the last part it looks like maybe u=\log{r\over a} is called for.

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outermeasure
 Post subject: Re: very difficult definite integral
PostPosted: Wed, 18 Apr 2012 04:56:21 UTC 
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Topic merged.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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