Series of product of legendre polynomials with shifted degree

Question

I am working on some quantum mechanics and I would love to find a closed expression for the series $$ S(x,y) = \sum_{l=0}^\infty P_l(x) P_{l+1}(y) $$ $x,y \in \left[ -1, 1\right]$ and also for its multisection $$ S_{qp}(x,y) = \sum_{l=0}^\infty P_{lq+p}(x) P_{lq+p+1}(y) $$ where $q \in \mathbb{N}$, $p \in \mathbb{N}_0$ and $p < q$ and $P_l$ is the $l$-th legendre polynomial.

The first series is a special case of a Hadamard product of two generating functions $F(x,t) \odot G(y,t)$ \begin{align} F(x,t) &= \frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{l=0}^\infty P_l(x) t^l\\ G(y,t) &= \frac{1}{t}\left(\frac{1}{\sqrt{1 - 2yt + t^2}} - 1 \right)= \sum_{l=0}^\infty P_{l+1}(y) t^l \; \text{.} \end{align} for $t=1$. Having the generating function for $F(x,t) \odot G(y,t)$ we would also have the multisection generating function, as shown here.

There is a way to compute the Hadamard product via an integral but I am not sure how (and whether) I can use this formula because it probably works only for $F$, $G$ being rational functions (a proof is here Derivation of the termwise Hadamard product of two generating functions.). I think the integral diverges in general.

I did some numerical calculations and it seems the expression shouldn't be so complicated. On Figure 1 is $S(x,y)$ with $x=1/2$ and $y$ being the $x$ axis. As $x$ moves, the point of discontinuity shifts, it is always at $y=x$.

Any ideas?

Answer 1

More generally, for $t\in \left[ -1,1\right]$,

$$\sum_{l=0}^\infty P_l(\cos\alpha) P_{l+1}(\cos\beta) t^l \!=\! \frac{2 \left(e^{-i \alpha } K\left(\frac{4 t \sin\alpha \sin\beta}{1-2 t\cos (\alpha +\beta )+t^2}\right)+\left(e^{-i \beta } t-e^{-i \alpha }\right) \Pi \left(\frac{2 i t \sin\beta }{e^{i \beta } t-e^{-i \alpha }}\bigg{|}\frac{4 t \sin\alpha \sin\beta}{1-2t\cos (\alpha +\beta)+t^2}\right)\right)}{\pi t \sqrt{1-2 t \cos (\alpha +\beta )+t^2}},$$

where $$K(m) = \int_0^{\pi/2} \frac{\mathrm{d}\varphi}{\sqrt{1-m \sin^2 \varphi}}, \qquad \Pi(n|m) = \int_0^{\pi/2} \frac{\mathrm{d}\varphi}{(1-n \sin^2 \varphi)\sqrt{1-m \sin^2 \varphi}}$$ are complete elliptic integrals of the first and the third kind, respectively.

Note that this result always turns out to be real. The first argument $n$ of $\Pi$ is complex, but there might be formulae for $\Pi$ expressing it in real arguments only.