# Infinite sum of a product of associated Legendre polynomias

I would like to ask you if any one of you have an idea how to tackle this problem: I have this sum

$$S=\sum_{\ell=1}^{\infty} \frac{2\ell+1}{\ell(\ell+1)}P_\ell^1(x)P_\ell^1(y)t^\ell$$

and I would like to know what is its exact form. Indeed I belive there is an exact form partially because of http://www-elsa.physik.uni-bonn.de/~dieckman/InfProd/InfProd.html#SeriesxofxLegendrexPolynomials where is similar sum

$$G(x,y,t)=\sum_{\ell=0}^{\infty} P_\ell(x) P_\ell(y) t^\ell$$

which equates

$$G=\frac{2}{\pi \sqrt{1+t^2-2txy+2t\sqrt{1-x^2}\sqrt{1-y^2}}}\mathrm{K}\left[\frac{4t\sqrt{1-x^2}\sqrt{1-y^2}}{1+t^2-2txy+2t\sqrt{1-x^2}\sqrt{1-y^2}}\right]$$

(with $\mathrm{K}\left[\,\cdot\,\right]$ being the elliptic integral of the first kind) and partially due to a huge amount of recurence formulas for Associated Legendre polynomials. For example I was able to rewrite my sum without denominators like

$$S = \frac{1}{\sqrt{1-x^2}\sqrt{1-y^2}}\sum_{\ell=1}^\infty (2\ell+1)\left(xP_\ell(x)-P_{\ell-1}(x)\right)\left(P_{\ell+1}(y)-yP_{\ell}(y)\right)$$

We have in total four summands, one of which is $(2\ell +1)P_\ell(x)P_\ell(y)$ which can be expressed simply as $2t\frac{\partial G}{\partial t}+G$, the other ones involve terms which sums to sums I have not been able to find on the internet. Maybe there is an prroach using just original sum, not expanding it, but I don't know...

EDIT:

Many approaches stuck on a problem of computing incomputable integrals, however, one can $G(x,y,t)$ write as a double integral across a unit surface with kernel $\frac{1}{|\vec{r}-\vec{r'}|}$ (as some potential from electrostatics or gravitational potential of circular object). Is there a possibility there is such a formula for my sum insted of a single integral?

• It seems to me that legibility is improved if instead of writing $$S=\sum_{l=1}^\infty \frac{2l+1}{l(l+1)}P_l^1(x)P_l^1(y)t^l,$$ one writes $$S=\sum_{\ell=1}^\infty \frac{2\ell+1}{\ell(\ell+1)}P_\ell^1(x)P_\ell^1(y)t^\ell.$$ – Michael Hardy Jun 15 '18 at 14:45
• @MichaelHardy Thank you, I have corrected it – Machinato Jun 15 '18 at 14:48

The summation can be written as \begin{align} S=&\sum_{\ell=1}^{\infty} \frac{2\ell+1}{\ell(\ell+1)}P_\ell^1(x)P_\ell^1(y)t^\ell\\ &=\sqrt{\left( 1-x^2 \right)\left( 1-y^2 \right)}\frac{\partial^2}{\partial x\partial y}\sum_{\ell=1}^{\infty}\left( \frac{1}{\ell}+\frac{1}{\ell+1} \right)P_\ell(x)P_\ell(y)t^\ell\\ &=\sqrt{\left( 1-x^2 \right)\left( 1-y^2 \right)}\frac{\partial^2}{\partial x\partial y}\left( T_0+T_1 \right) \end{align} where, for $k=0,1$, $$T_k=\sum_{\ell=1}^{\infty}P_\ell(x)P_\ell(y)\frac{t^\ell}{\ell+k}$$ Using the given identity, we have \begin{align} G(x,y,\tau)&=\sum_{\ell=0}^{\infty} P_\ell(x) P_\ell(y) \tau^\ell\\ \frac{1}{t}\int_0^\tau G(x,y,\tau)\,d\tau&=\sum_{\ell=0}^{\infty} P_\ell(x) P_\ell(y) \frac{t^\ell}{\ell+1}\\ &=T_1+1\\ \int_0^t\frac{G(x,y,\tau)-G(x,y,0)}{\tau}\,d\tau&=\sum_{\ell=1}^{\infty} P_\ell(x) P_\ell(y) \frac{t^\ell}{\ell}\\ &=T_0 \end{align} with $G(x,y,0)=1$, one obtains finally $$S=\sqrt{\left( 1-x^2 \right)\left( 1-y^2 \right)}\frac{\partial^2}{\partial x\partial y}\int_0^t\left[ \frac{G(x,y,\tau)-1}{\tau}+\frac{G(x,y,\tau)}{t}\right]\,d\tau$$ To go further by this approach seems difficult as it requires the calculation of likely complicated integrals.
• Very nice approach indeed, however, my approach also stucked on a similar problem, computing incomputable integrals, I have seen, however, one can $G(x,y,t)$ write as a double integral across some surface with kernel $\frac{1}{|\vec{r}-\vec{r'}|}$. Is there a possibility there is such a formula? I will add this into the original question ... – Machinato Jun 15 '18 at 18:48