# Series of product of legendre polynomials with shifted degree

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?

• Interesting problem! Indeed the graph of $S$ looks like it admits quite some structure; the simple mathematica code f[x_, y_, nmax_] := Sum[LegendreP[n, x]*LegendreP[n + 1, y], {n, 0, nmax}] Plot3D[f[x, y, 600], {x, -1, 1}, {y, -1, 1}] produced the following images: 1 2 3. While I think the bends at $(-1,1),(1,-1)$ are truncation artifacts, I'm not so sure the same is true at $(-1,-1),(1,1)$ Commented May 9, 2023 at 6:19
• @FrederikvomEnde For $y=1$ the function is just $S(x,1) = 1/ \sqrt{2 - 2x}$ (it's the F generating function) and we see that the bends at $(-1,1)$ and $(1,-1)$ are indeed just a numerical error, the real value there is $1/2$ at $(-1,1)$ and $-1/2$ at $(1,-1)$. From the same reason the value at $(1,1)$ is diverging to $+ \infty$ and value at $(-1-1)$ to $- \infty$ (because $P_l(-1) P_{l+1}(-1) = -1$) . I am not sure whether all the values at $x=y$ are diverging but my guess is yes. Commented May 9, 2023 at 11:05

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.