1
$\begingroup$

I have encountered the following sum involving Legendre polynomials, which I hope to reduce to something involving a $\delta$-function: $$ \frac{d^2}{dx^2} \sum_{\ell=0}^{\infty} \frac{2 \ell + 1}{2 \ell (\ell+1)} P_{\ell}(x) P_{\ell}(y) $$

I have played around with the recurrence relation for derivatives of Legendre polynomials without much success ($(1-x^2) P_{\ell}'(x) = -\ell x P_{\ell}(x) + \ell P_{\ell-1}(x)$). I have done some numerical investigation to try and gain intuition, and it does appear that the result vanishes for $x \neq y$, as expected for something proportional to $\delta(x-y)$. However, I've run into tricky divergences and other problems for the $x=y$ case.

If anyone has come across a sum or identity similar to this, I would be most obliged.

$\endgroup$

1 Answer 1

1
$\begingroup$

Firstly, I assume that $l$ goes from $1$ to infinity, otherwise you are dividing by $0.$ Secondly, without the derivative, mathematica says that $\sum_{l=1}^\infty \frac{2l+1}{l(l+1)} P_l(x) P_l(y) = $

if $x < y,$ $ -\log (-(x-1) (y+1))+2 \log (2)-1$,

if $x > y$ then $ -\log (-(x+1) (y-1))+2 \log (2)-1$

and $0$ otherwise.

You can differentiate this twice and see what you get.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .