# Integral over triple Legendre polynomials involving derivatives

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?

## 1 Answer

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}