# Closed form expression for series involving Legendre polynomials

Given $$-1 \leq x \leq 1$$ and $$0 \leq \eta \leq 1$$, I am interested in computing $$E(x,\eta) = \sum_{\ell = 0}^{+ \infty} |P_{\ell} (0)|^{2} \, P_{\ell} (x) \, \eta^{\ell} ,$$ with $$P_{\ell}$$ the usual Legendre polynomials.

Can one find a closed form expression for this function?

NB1: Should this be useful, $$E(x,\eta)$$ is, in essence, the gravitational potential between two massive circles centered around the same location, inclined by a respective angle $$\cos^{-1}(x)$$ and with a ratio of radii given by $$\eta$$.

NB2: Of course, in the limit $$x \to 1$$ or $$\eta \to 0$$, I expect that I could perform appropriate limited developments.

As $$P_{2n+1}(0)=0$$ and $$P_{2n}(0)=(-1)^n\frac{(1/2)_n}{n!}$$, the proposed series reads \begin{align} E(x,\eta) &= \sum_{\ell = 0}^{+ \infty} |P_{\ell} (0)|^{2} \, P_{\ell} (x) \, \eta^{\ell} \\ &=\sum_{n=0}^\infty \frac{\left(\frac{1}{2}\right)_{n}^{2} }{n !^{2}}P_{2n}(x)\eta^{2n} \end{align}

Interestingly, such a series was discussed in connection with Ramanujan type series for $$1/\pi$$ in this paper (see also here). Wan and Zudilin prove the following identity \begin{align} \sum_{n=0}^\infty\frac{\left(\frac{1}{2}\right)_{n}^{2} }{n !^{2}}\,&P_{2n} \left( \frac{\left(X +Y \right) \left(1-X Y\right)}{\left(X -Y \right) \left(1+X Y \right)}\right) \left(\frac{X -Y}{X Y +1}\right)^{2 n}\\ &=\frac{\left(1+X Y\right)}{2} \,_2F_1\left(\frac{1}{2},\frac{1}{2};1;1-X^{2}\right)\,_2F_1\left(\frac{1}{2},\frac{1}{2};1;1-Y^{2}\right) \end{align} the domain of validity for the parameters $$X,Y$$ is relatively complex, but numerical experiments seem to indicate that most of the values of $$-1\le x\le 1$$ and $$0\le \eta\le1$$ (if not all) could be covered.

• Your formula and mine combined means a weighted integral of an elliptic function can be written as a product of elliptic functions, a pleasant surprise. Commented Jan 29, 2022 at 18:42
• Yes, that is a nice connection, even if the transformations of the involved parameters are rather complicated! Commented Jan 29, 2022 at 20:30

A single integral form of the series in question is $$\sum_{n=0}^\infty t^n P_n(0)^2 P_n(\cos{a}) = \frac{4}{\pi^2}\int_0^\pi \frac{db}{ u_+(b)+u_-(b)}K\big(\frac{(u_+(b)-u_-(b))^2}{(u_+(b)+u_-(b))^2}\big)$$ where $$u_+(b) = \sqrt{1-2t\cos{(a+b)} + t^2} \quad,\quad u_-(b) = \sqrt{1-2t\cos{(a-b)} + t^2}$$ and the fact that $$u_\pm$$ also depends on $$a$$ and $$t$$ has been suppressed.

$$K(x)$$ is the complete elliptic function of the first kind, implemented in Mathematica as $$K(x) =$$EllipticK[x].

The derivation follows from integrating over $$b$$

$$\sum_{n=0}^\infty t^n P_n(\cos{a}) P_n(\cos{b}) =\frac{4}{\pi} \frac{1}{ u_+(b)+u_-(b)}K\big(\frac{(u_+(b)-u_-(b))^2}{(u_+(b)+u_-(b))^2}\big)$$

as I found in the answer to An infinite series involving Legendre polynomials Note that the argument of K in my answer is the square of the one in the reference, and that is to make it consistent with Mathematica's input. (There is a confusing array of notations for elliptic integrals.)

Gradshteyn and Rhyzhik 7.226.1, symmetry of the integrand, and the known closed-value for $$P_n(0)$$ can be manipulated to read $$\int_0^\pi P_n(\cos{b}) db = \pi P_n(0)^2$$ Hence, the derivation is easy, once the right formulas are available. The answer has been checked numerically for many $$a$$ and $$t.$$ It is highly unlikely there is a simpler form.