While investigating a problem in acoustic scattering in bounded domains, I encountered the following integral: $$\int_{-1}^{1}\frac{\text{P}_n(x)\text{P}_m(x)}{\sqrt{1-x^2}}\mathrm{d}x$$ Where $\text{P}_n(x)$ is the nth order Legendre polynomial of the first kind. Is there a neat closed-form solution (or approximation) to this integral?

I know one can replace each Legendre polynomial with a finite sum of Chebyshev polynomials using Gegenbauer function, but that still produces a lengthy sum of fractions which is too cumbersome to be used in further calculations.

Thank you for your help!


1 Answer 1


Here is a single sum representation. I doubt it simplifies further:

$$ S_{n\,m}:=\int_{-1}^1 \frac{P_n(x)\,P_m(x)}{\sqrt{1-x^2}} dx = \pi \, \frac{ 1 + (-1)^{n+m}}{2} \cdot $$ $$ \cdot \sum_{r=0}^m \frac{A_{m-r}\,A_r\,A_{n-r}}{A_{m+n-r}} \, \frac{2m+2n-4r+1}{2m+2n-2r+1} \Big( 2^{-(m+n-2r)}\binom{m+n-2r}{(m+n)/2 - r} \Big)^2 $$ where $$A_m := 2^{-m}\binom{2m}{m} $$

The expression is derived from two formulas. The first is the Neumann-Adams formula for the linearization of the product of two Legendre polynomials,

$$ P_n(x)\,P_m(x) = \sum_{r=0}^m \frac{A_{m-r}\,A_r\,A_{n-r}}{A_{m+n-r}} \, \frac{2m+2n-4r+1}{2m+2n-2r+1} P_{m+n-2r}(x) $$

The second formula is from the integral (see Gradstheyn 7.132.1) $$ \int_{-1}^1 \frac{P_q(x)}{\sqrt{1-x^2}} dx = \pi \, \frac{1+(-1)^{q}}{2} \Big( 2^{-q} \binom{q}{q/2} \Big)^2$$


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.