# Legendre Polynomial Integral

How can I evaluate $$\int_{-1}^1 P_n(x)P_l(x)x^k dx$$ when $k$ is even?

Or what might be a source where I could find integrals like this?

The integral $$\int\limits_0^1 x^k P_m(x)P_n(x)dx$$ is evaluated in terms of the hypergeometric function $_3F_2$ in http://link.springer.com/article/10.1007/BF01650571 (Some integrals containing products of legendre polynomials, by L. Carlitz). The evaluation is based on $$P_m(x)P_n(x)=\sum\limits_{r=0}^{\mathrm{min}(m,n)}\frac{A_rA_{m-r}A_{n-r}}{A_{m+n-r}}\,\frac{2m+2n-4r+1}{2m+2n-2r+1}P_{m+n-2r}(x),$$ with $$A_r=\frac{(2r-1)!!}{2^rr!},$$ and $$\int\limits_0^1 x^k P_n(x)dx=\frac{k(k-1)\cdots(k-n+2)}{(k+n+1)(k+n-1)\cdots(k-n+3)}.$$ The following formulas $$\int\limits_{-1}^1 x P_m(x)P_n(x)dx=\left\{\begin{array}{cc}\frac{2(m+1)}{(2m+1)(2m+3)}, & n=m+1, \\ \frac{2m}{(2m-1)(2m+1)}, & n=m-1,\end{array}\right .$$ and $$\int\limits_{-1}^1 x^2 P_m(x)P_n(x)dx=\left\{\begin{array}{cc}\frac{2(m+1)(m+2)}{(2m+1)(2m+3)(2m+5)}, & n=m+2, \\ \frac{2(2m^2+2m-1)}{(2m-1)(2m+1)(2m+3)}, & n=m, \\ \frac{2m(m-1)}{(2m-3)(2m-1)(2m+1)}, & n=m-2,\end{array}\right .$$ are given as an exercise 12.9.2 (p. 806) in the book Mathematical Methods for Physicists, 6th Edition, by G.B Arfken and H.J. Weber: https://www.amazon.com/Mathematical-Methods-Physicists-George-Arfken/dp/0120598760
• I think $A_r$ is supposed to be $\frac{(2r-1)!!}{2^rr!}$ though? Jul 14, 2016 at 1:14
The integral can be evaluate from the formula which express the product of two Legendre polynomials as a sum of Legendre polynomials (proved by F. Neumann and J.C. Adams, 1878), see e.g. the paper of W.A. Salam, On the product of two Legendre polynomials (http://www.mscand.dk/article/viewFile/10471/8492) and then the formulas which evaluate $$\int_0^1 x^k P_m(x)dx,$$ in terms of product of Gamma functions, see formulas 22.13.8 and 22.13.9 p.786 of Abramowitz and Stegun.