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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!

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1 Answer 1

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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) $$ \int_{-1}^1 \frac{P_q(x)}{\sqrt{1-x^2}} dx = \pi \, \frac{1+(-1)^{q}}{2} \Big( 2^{-q} \binom{q}{q/2} \Big)^2$$

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