0
$\begingroup$

I know the integral over the triple product of Legendre polynomials (see Legendre Polynomials Triple Product), which reads

\begin{align} \int_{-1}^{1} P_k(x)\,P_l(x)\, P_m(x) \;\mathrm{d}x = 2 \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2 \end{align}

where the big parenthesis is Wigner-3$j$ symbol.

But I encountered a similar integral in a physics problem \begin{align} I=\int_{-1}^{1} P_k'(x)\,P_l'(x)\, P_m(x) (1-x^2)\;\mathrm{d}x \end{align} where the prime $'$ means the derivative with respect to $x$. I don't know is there a similar closed form solution of the above integral?

I tried using the recurrence relation \begin{equation} (1-x^2)P_n'(x)=(n+1)[xP_n(x)-P_{n+1}(x)] \end{equation} and the integral becomes \begin{equation} I=(k+1)(l+1)\int_{-1}^1 [xP_k(x)-P_{k+1}(x)][xP_{l}(x)-P_{l+1}(x)]P_m(x) \frac{1}{1-x^2}\;\mathrm{d} x \end{equation} It seems not helpful.

Any ideas?

$\endgroup$
0
$\begingroup$

I found there indeed exists a closed form solution which reads

\begin{align} I=\int_{-1}^{1} P_k'(x)\,P_l'(x)\, P_m(x) (1-x^2)\;\mathrm{d}x = [k(k+1)+l(l+1)-n(n+1)] \begin{pmatrix} k & m & l \\ 0 & 0 & 0 \end{pmatrix}^2 \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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