Closed form expression for series involving Legendre polynomials

Question

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.

Answer 1

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.

Answer 2

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.