Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I encountered the following integral in my (physics) research, and I've yet to find an analytic solution: $$I(n_1,n_2,n_3) = \int_{-1}^{1} d(\cos\theta_1) \int_{-1}^{1} d(\cos\theta_2) P_{n_1}(\cos\theta_1) P_{n_2}[\cos(\theta_1-\theta_2)] P_{n_3}(\cos\theta_2)$$ where $P_n(x)$ is the nth Legendre polynomial. For reference, the orthogonal polynomial relation for Legendre polynomials is: $$\int_{-1}^{1} d(\cos\theta) P_{n_1}(\cos\theta) P_{n_2}(\cos\theta) = \int_{0}^{\pi} d\theta (\sin\theta) P_{n_1}(\cos\theta) P_{n_2}(\cos\theta) = \frac{2}{2n_1+1}\delta_{n_1,n_2}$$

Mathematica does not have much trouble in evaluating the integral exactly (Integrate, not NIntegrate) for a given $\{n_1, n_2, n_3\}$.

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 $I(n_1, n_2, n_3)$ is not diagonal. Even the diagonal terms contain factors which I've yet to figure out, here's what I see so far: $$I(n,n,n) = 2^n \left[\frac{2}{2n+1}\right]^2 \times \left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{5},\frac {4}{35},\frac{4}{63},\frac{8}{231},\frac{8}{429}, \frac{64}{6435}, ... \right\}$$ where the bracketed terms correspond to $n=0,1,2,...$

Many of the off-diagonal terms appear to contain a $\pi^2$ term and some power of 2.

Do you have any suggestions? Useful variable transformations, pattern recognition? Thanks in advance.

share|improve this question

2 Answers 2

I guess this is one of these cases where the spherical harmonics addition theorem $$ P_l(\cos \gamma) = \frac{4\pi}{2l+1} \sum_{m=-l}^l Y^*_{lm}(\theta_1,\phi_1) Y_{lm}(\theta_2,\phi_2)$$ comes in handy; here, $\gamma$ is the angle between two vectors having spherical coordinates $(\theta_1,\phi_1)$ and $(\theta_2,\phi_2)$ and $Y_{lm}$ are the spherical harmonics.

In your case, we choose $\phi_1=\phi_2=0$ and the fact that $$Y_{lm}(\theta,0)= \sqrt{\frac{(2l+1)(l-m)!}{4\pi(l+m)!}} P_l^m(\cos\theta) $$ with $P_l^m$ the associated Legendre polynomials. The addition theorem thus reads $$P_l [\cos(\theta_1 - \theta_2)]= \sum_{m=-l}^l \frac{(l-m)!}{(l+m)!}P_l^m(\cos\theta_1)P_l^m(\cos\theta_2).$$

Plugging the addition theorem into your integral yields, $$I(n_1,n_2,n_3) = \int_{-1}^{1}\! dx_1 \int_{-1}^{1}\! dx_2\, P_{n_1}(x_1) \left[\sum_{m=-n_2}^{n_2} \frac{(n_2-m)!}{(n_2+m)!}P_{n_2}^m(x_1)P_{n_2}^m(x_2) \right] P_{n_3}(x_2).$$

In a last step, we need to know the integral $$p(k,l,m)=\int_{-1}^1 \!dx \,P_k(x) P_l^m(x)$$ in terms of which we can find the expression $$I(n_1,n_2,n_3) = \sum_{m=-n_2}^{n_2} \frac{(n_2-m)!}{(n_2+m)!} p(n_1,n_2,m) p(n_3,n_2,m).$$

It turns out that the general expression for $p(k,l,m)$ is rather complicated. However, for $k=l$, we have the simple expression $$p(l,l,m)= \begin{cases} \frac{2(-1)^{m/2} l! (2l)!}{(1+2l)! (l-m)!} & m\text{ even},\\ 0& m\text{ odd}.\end{cases} $$

For the diagonal elements, we obtain $$I(n,n,n)= \sum_{\substack{m=-n;\\m\text{ even}}}^n \frac{1}{(n+m)!}\frac{4 \,n!^2\, (2 n)!^2}{(2 n+1)!^2\, (n-m)!} . $$

As the expression for the diagonal element is already rather complicated, and I am afraid the off diagonal elements will be even more cumbersome. However, using the paper linked above, you should be able to obtain an explicit expression.

share|improve this answer
up vote 2 down vote accepted

This solution is a continuation of Fabian's answer. Thanks for pointing me in the right direction!

Starting from the original definition of $I(n_1,n_2,n_3)$: $$I(n_1,n_2,n_3) = \int_{-1}^{1} d(\cos\theta_1) \int_{-1}^{1} d(\cos\theta_2) P_{n_1}(\cos\theta_1) P_{n_2}[\cos(\theta_1-\theta_2)] P_{n_3}(\cos\theta_2)$$ we can expand the middle Legendre polynomial in terms of associated legendre polynomials as follows: $$P_l [\cos(\theta_1 - \theta_2)]= \sum_{m=-l}^l \frac{(l-m)!}{(l+m)!}P_l^m(\cos\theta_1)P_l^m(\cos\theta_2)$$ Then $I(n_1,n_2,n_3)$ is written: $$I(n_1,n_2,n_3) = \int_{-1}^{1} dx_1 \int_{-1}^{1} dx_2\, P_{n_1}(x_1) \left[\sum_{m=-n_2}^{n_2} \frac{(n_2-m)!}{(n_2+m)!}P_{n_2}^m(x_1)P_{n_2}^m(x_2) \right] P_{n_3}(x_2)$$ Defining the overlap of a Legendre polynomial with an associated Legendre polynomial as $$X(n,n_2,m_2)=\int_{-1}^1 dx\, P_{n}(x) P_{n_2}^{m_2}(x)$$ so that $$I(n_1,n_2,n_3) = \sum_{m_2=-n_2}^{n_2} \frac{(n_2-m)!}{(n_2+m)!} X(n_1,n_2,m_2) X(n_3,n_2,m_2)$$

A single sum expression has been derived for the overlap integral of two associated Legendre polynomials in J. Phys. A: Math. Gen. 32 (1999) 2601-2603 (along with a small correction in a comment paper: J. Phys. A: Math. Gen. 35 (2002) 4187-4188). The solution in that paper is for $$ Z(n_1,m_1,n_2,m_2) = \int_{-1}^1 dx\, P_{n_1}^{m_1}(x) P_{n_2}^{m_2}(x) $$ but we only need $X(n,n_2,m_2)=Z(n,0,n_2,m_2)$. This simplifies the resulting expressions slightly. When $m_2=0$ we have the overlap of two Legendre polynomials, so $$X(n,n_2,0) = \frac{2}{2n+1} \delta_{n,n_2}$$ When $m_2 \neq 0$, $$X(n,n_2,m_2 \neq 0) = A(n_2,m_2) \sum_{l=l_{\text{min}}}^{l_{\text{max}}} D(l,m_2)(2l+1) \begin{pmatrix} n & n_2 & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} n & n_2 & l \\ 0 & m_2 & -m_2 \end{pmatrix} $$ where $ \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} \text{ is the Wigner 3-j symbol} $ and \begin{aligned} A(n,m) =& (-1)^{\tau(m)} |m|\, 2^{|m|-2} \sqrt{\frac{(n+m)!}{(n-m)!}} \\ \tau(m) =& \begin{cases}0 & \text{if } m \geq 0 \\ m & \text{if } m < 0\end{cases} \\ D(l,m) =& \begin{cases} \left(1+(-1)^{l+|m|}\right) \sqrt{\frac{(l-|m|)!}{(l+|m|)!}} \frac{\Gamma(l/2)\Gamma[(l+|m|+1)/2]}{\Gamma[(l+3)/2][(l-|m|)/2]!} & \text{if } l+m \text{ is even} \\ 0 & \text{if } l+m \text{ is odd} \end{cases} \\ l_{min} =& \begin{cases}|n-n_2| & \text{if } |n-n_2| \geq |m_2| \\ |m_2| & \text{if } |n-n_2| < |m_2|\end{cases} \\ l_{max} =& n + n_2 \end{aligned}

This completes the general solution for $I(n_1,n_2,n_3)$. As pointed out by Fabian, the diagonal elements have a considerably simpler form: $$X(n,n,m) = \begin{cases} \frac{2(-1)^{m/2} n! (2n)!}{(2n+1)! (n-m)!} & m\text{ even}\\ 0& m\text{ odd}\end{cases}$$ so that $I(n,n,n)$ can be quickly calculated from $$I(n,n,n) = \sum_{\substack{m=-n\\m\text{ even}}}^{n} \frac{1}{(n+m)!}\frac{4[n!]^2[(2n)!]^2}{[(2n+1)!]^2(n-m)!}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.