- #1

alfredska

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## Homework Statement

I encountered the following integral in my research, and I've yet to find an analytic solution:

[tex]I(n_1,n_2,n_3) = \int_{-1}^{1} d(\cos\theta_1) \int_{-1}^{1} d(\cos\theta_2) P_{n_1}(\cos\theta_1) P_{n_2}[\cos(\theta_1-\theta_2)] P_{n_3}(\cos\theta_2)[/tex]

where [itex]P_n(x)[/itex] is the n

^{th}Legendre polynomial.

## Homework Equations

Orthogonal polynomial relation:

[tex]\int_{-1}^{1} d(\cos\theta) P_{n_1}(\cos\theta) P_{n_2}(\cos\theta) = \int_{0}^{\pi} d\theta (\sin\theta) P_{n_1}(\cos\theta) P_{n_2}(\cos\theta) = \frac{2}{2n_1+1}\delta_{n_1,n_2}[/tex]

## The Attempt at a Solution

Mathematica does not have much trouble in evaluating the integral exactly (Integrate, not NIntegrate) for a given set of [itex]\{n_1, n_2, n_3\}[/itex].

Code:

`k[n1_,n2_,n3_]:=Integrate[Sin[x]Sin[y]LegendreP[n1,Cos[x]]LegendreP[n2,Cos[x-y]]LegendreP[n3,Cos[y]],{x,0,Pi},{y,0,Pi}]`

I can see already that [itex]I(n_1, n_2, n_3)[/itex] is not diagonal.

The diagonal terms contain the factor [tex]\frac{2^n}{(2n+1)^2}[/tex]

but that is not all. They also contain something that looks like a truncated double factorial: [itex](2n-1)![/itex], where the truncation point is not clear yet. And still there's more to the series that I haven't figured out:

[tex]I(n,n,n) = 2^n \left[\frac{2}{2n+1}\right]^2 \times \left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{5},\frac{4}{35},\frac{4}{63},\frac{8}{231},\frac{8}{429}, \frac{64}{6435}, ... \right\}[/tex]

for [itex]n=0,1,2,...[/itex]

The off-diagonal terms appear to all contain a [itex]\pi^2[/itex] term, and some power of 2.

I suspect that the solution will need to be written in the form

[tex]I(n_1,n_2,n_3) = \delta_{n_1,n_2}\delta_{n_2,n_3}f(n_1) + (1-\delta_{n_1,n_2}\delta_{n_2,n_3})\pi^2 g(n_1,n_2,n_3)[/tex]

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