3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $$ S=\sum_{\ell=1}^\infty \frac{2\ell+1}{\ell(\ell+1)}P_\ell^1(x)P_\ell^1(y)t^\ell. $$ $\endgroup$ – Michael Hardy Jun 15 '18 at 14:45
  • 1
    $\begingroup$ @MichaelHardy Thank you, I have corrected it $\endgroup$ – Machinato Jun 15 '18 at 14:48
2
$\begingroup$

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 $k=0,1$, \begin{equation} T_k=\sum_{\ell=1}^{\infty}P_\ell(x)P_\ell(y)\frac{t^\ell}{\ell+k} \end{equation} 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 \begin{equation} 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 \end{equation} To go further by this approach seems difficult as it requires the calculation of likely complicated integrals.

$\endgroup$
  • $\begingroup$ 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 ... $\endgroup$ – Machinato Jun 15 '18 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.