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I have to solve the following integral:

\begin{align} \int_{-1}^{1} \left(x^2 -1\right)^3 P_k(x)\,P_l(x)\, P_m(x) \;dx \end{align} where $P_{k,l,m}$ are Legendre Polynomials

The triple product \begin{align} \int_{-1}^{1} P_k(x)\,P_l(x)\, P_m(x) \;dx = 2 \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2 \end{align} using the special case of $3j$ symbol form \begin{align} \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix} &= (-1)^s \sqrt{(2s-2k)! (2s-2l)! (2s-2m)! \over (2s+1)!} {s! \over (s-k)! (s-l)! (s-m)!} \\ & \mbox{for $2s=k+l+m$ even} \\[3pt] \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix} &= 0 \quad\mbox{for $2s=k+l+m$ odd} \\ \end{align}

I'm sure you should be able to solve this by doing integration by parts but can't seem to get it to work. Any tips?

So using the answer below I think you get the following for step 1 of 3

\begin{multline} \int_{-1}^{1}(x^2-1)^3 P_k P_l P_m = \overbrace{(x^2-1)^3\frac{(P_{k+1} - P_{k-1})}{2k+1} P_l P_m \Big]_{-1}^1}^\text{ = 0}\\ -\int_{-1}^{1} \frac{(P_{k+1} - P_{k-1})}{2k+1}(x^2-1)^2\Big( 6xP_l P_m \\ + (1+l) P_m(P_{l+1} - P_{l-1}) + (1+m) P_l(P_{m+1} - P_{m-1}) \Big) \; dx\\ \end{multline}

Not sure if the formula for integration works as I think the $6xP_lP_m$ term might cause problems?

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    $\begingroup$ Has anyone seen discussion of this kind of problem for four polynomials in the integrand? i.e. \int_{-1}^1 P_m(x)P_n(x)P_l(x)P_p(x) dx $\endgroup$ Oct 16 '18 at 18:23
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I think the best way to approach this is as follows, note that \begin{align} (x^2 -1 ) = \frac{P_2 - 2}{3} \end{align} You can then use the following definition \begin{align} P_kP_l = \sum_{m=|k-l|}^{k+l} \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1)P_m \end{align} This allows the integral to be written as follows \begin{align} \int_{-1}^{1} (x^2-1)^3P_iP_jP_k \; dx &= \int_{-1}^{1} \frac{1}{9}\left(P_2^3 + . . .-8 \right) P_i P_j P_k \; dx \end{align} The most difficult term to deal with is the $ P_2^3 P_i P_j P_k$ \begin{align} P_2^3 P_i P_j P_k &= \sum_{m=0}^{4} \begin{pmatrix} 2 & 2 & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1)P_m P_2 P_i P_j P_k \\ &= \sum_{m=0}^{4} \begin{pmatrix} 2 & 2 & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1) \sum_{n=|m-2|}^{m+2} \begin{pmatrix} 2 & m & n \\ 0 & 0 & 0 \end{pmatrix}^2 (2n+1)P_n P_i P_j P_k \\ &= \sum_{m=0}^{4} \begin{pmatrix} 2 & 2 & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1) \sum_{n=|m-2|}^{m+2} \begin{pmatrix} 2 & m & n \\ 0 & 0 & 0 \end{pmatrix}^2 (2n+1) \sum_{l=|n-i|}^{n+i} \begin{pmatrix} n & i & l \\ 0 & 0 & 0 \end{pmatrix}^2 (2l+1) P_l P_j P_k \end{align} Which can then make use of the usual triple integral. All other terms can be solved for in a similar manner.

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  • $\begingroup$ Hi! I think this approach may help me too: do you know of any analogous definitions for order 1 Legendre polynomials to reduce a product of two into a linear sum? It would be really useful right now! $\endgroup$ Dec 8 '18 at 16:26
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You could integrate one of the $P_k(x)$ and take the derivative of the rest. The power of $(1-x^2)$ gets reduce by the fact that $$\partial_x P_l(x) = \frac{(1+l) [ P_{l+1}(x)-x P_l(x) ]}{x^2-1}.$$ You have to apply partial integration a few times and you will generate a bunch of Legendre triple product.

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  • $\begingroup$ Think this way will work, thanks for the help. $\endgroup$
    – Matt
    Dec 21 '12 at 2:18
  • $\begingroup$ Just wondering, where did you get the formula above? $\endgroup$
    – Matt
    Dec 21 '12 at 15:30
  • $\begingroup$ @Matt: I hope it is correct. I simply looked it up here. $\endgroup$
    – Fabian
    Dec 21 '12 at 16:00
  • $\begingroup$ thanks for info, I think it might be wrong, wikipedia has a different one (which I think is also wrong) as I worked out the following: \begin{align} P'_n = \frac{n(n+1)}{(1-x^2)(2n+1)} \left[P_{n+1} - P_{n-1} +c \right] \end{align} Which works when I do the differentiation. $\endgroup$
    – Matt
    Dec 21 '12 at 18:07
  • $\begingroup$ got it from the following definition: \begin{align} \frac{d}{dx} \left[ (1-x^2) \frac{d}{ dx} P_n(x) \right] + n(n+1)P_n(x) = 0 \end{align} $\endgroup$
    – Matt
    Dec 21 '12 at 18:16

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