# Generating function of the product of Legendre polynomials

The generating function of the product of Legendre polynomials for the same $$n$$ is given by

\begin{aligned} \sum_{n=0}^{\infty} z^{n} \mathrm{P}_{n}(\cos \alpha) \mathrm{P}_{n}(\cos \beta)&=\frac{\mathrm{F}\left(\frac{1}{2} ,\frac{1}{2} ;1;\frac{4z\sin \alpha \sin \beta }{1-2z\cos (\alpha +\beta )+z^{2}}\right)}{\sqrt{1-2z\cos (\alpha +\beta )+z^{2}}} \\&=\frac{2K\left(\sqrt{\frac{4z\sin \alpha \sin \beta }{1-2z\cos (\alpha +\beta )+z^{2}}}\right)}{\pi \sqrt{1-2z\cos (\alpha +\beta )+z^{2}}} \end{aligned} where $$P_n$$ is a Legendre polynomial, $$F$$ is a hypergeometric function and $$K$$ is a complete elliptic integral of the first kind. https://www.researchgate.net/publication/269015726_A_generating_function_for_the_product_of_two_Legendre_polynomials

However, in the study of quantum physics, we need the similar result as above for following expression: \begin{align} \sum_{n=0}^{\infty} z^{n} \mathrm{P}_{n}(x) \mathrm{P}_{n-1}(x) \end{align}

Is there any result known? Any hints for this problem would be appreciated.

• see this post of mine math.stackexchange.com/a/4699090/240067, when I have time, I can provide with the derivation. It's relatively straightforward Commented May 14, 2023 at 17:12

Using the recursion relation $$P_{n-1} (x) = x P_n (x) - \frac{x^2 -1}{n} \frac{d}{dx} P_n (x) \ ,$$ you can reduce your expression to a sum of the generating function you quote and a combined derivative (the $$d/dx$$) and integral w.r.t. $$z$$ (to generate the $$1/n$$) thereof.