# Closed form of sum of infinite series of Legendre polynomials

I was working on a research and we end up to discover that the Green's function on some domain is of the form :$$G = \frac{-1}{4\pi}\sum_{l=0}^{\infty}\frac{2l+1}{l(l+1)+\frac{1}{\alpha}}P_l(x)$$ where $\alpha$ $\in$ $\mathbb{R}^+$. I am seeking the closed form of this expression, is there any hints ? Thank you

• For $\alpha=4$, Mathematica can compute the result as $-\frac{1}{2\pi}K(\frac{1+x}{2})$ where $K(m)=\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m \sin^2{\theta}}}$ is the complete elliptic integral of the first kind. If nothing else, that gives a reference point for a perturbative expansion. Commented Apr 25, 2017 at 16:56
• @Semiclassical I did not understand : that gives a reference point for a perturbative expansion , can you please explain more? thanks Commented Apr 25, 2017 at 17:02
• Consider your function as $G(x,\alpha)$. Then you can write $$G(x,\alpha)=G(x,4)+c_1 (\alpha-4)+c_2(\alpha-4)^2+\cdots$$ i.e. a Taylor series expansion about $\alpha=4$. Since $G(x,4)$ can be computed exactly (as above) you can in principle approximate $G(x,\alpha)$ in the nbhd of this point by truncating this expansion. (This may not be much help, though, since $c_1$ doesn't have a nice closed-form as best I can tell...) Commented Apr 25, 2017 at 17:05
• It would be useful to know the origin of the term $\frac{2l+1}{l(l+1)+\frac{1}{\alpha^2}}$ since the most straightforward approach is to consider the generating function for Legendre polynomials and apply to it the linear operator bringing $z^l$ into $\frac{2l+1}{l(l+1)+\frac{1}{\alpha^2}}$. Commented Apr 25, 2017 at 18:13
• The case $\alpha=2$, leading to a complete elliptic integral, separates the cases for which $l(l+1)+\frac{1}{\alpha^2}$ has real roots from the cases it only has complex roots. That should affect the "oscillating behaviour" of $G(\alpha)$. Commented Apr 25, 2017 at 18:15

We have $$\frac{1}{\sqrt{1-2xt+t^2}}=\sum_{l\geq 0} t^l P_l(x) \tag{1}$$ and if for any $\alpha>0$ we set $\beta_{\pm} = \frac{-\alpha\pm\sqrt{\alpha(\alpha-4)}}{2\alpha}$ we have $$\frac{2l+1}{l(l+1)+\frac{1}{\alpha}}=\frac{1}{l-\beta_+}+\frac{1}{l-\beta_-} \tag{2}$$ so: $$\sum_{l\geq 0}\frac{2l+1}{l(l+1)+\frac{1}{\alpha}}P_l(x) = \int_{0}^{1}\frac{dt}{t\sqrt{1-2xt+t^2}}\left(\frac{1}{t^{\beta_+}}+\frac{1}{t^{\beta_-}}\right)\,dt \tag{3}$$ and the RHS of $(3)$ can be estimated by studying the behaviour of the integrand function at its singular points $t=0$ and $t=x\pm\sqrt{x^2-1}$. If $\alpha=4$ we also have an explicit closed form depending on a complete elliptic integral of the first kind.