What's the Legendre of zero? I know that this question may look like meaningless, but through solving a question, I encountered with this form of Legendre.
$ P_{l}(0)$ or a sum over it. You have any idea about it?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Use the generating function, with $\ds{\verts{h}\ <\ 1}$,
$\ds{{1 \over \root{1 - 2xh + h^{2}}}
     =\sum_{\ell\ =\ 0}^{\infty}h^{\ell}\,{\rm P}_{\ell}\pars{x}}$ such that:

\begin{align}
\sum_{\ell\ =\ 0}^{\infty}h^{\ell}\,{\rm P}_{\ell}\pars{0}
&={1 \over \root{1 + h^{2}}}=\sum_{\ell\ =\ 0}^{\infty}{-1/2 \choose \ell}h^{2\ell}
=\sum_{\ell\ =\ 0}^{\infty}{\ell - 1/2 \choose \ell}\pars{-1}^{\ell}h^{2\ell}
\\[5mm]&=\sum_{{\vphantom{\LARGE A}\ell\ =\ 0 }\atop \ell\ \mbox{even}}^{\infty}
{\ell/2 - 1/2 \choose \ell/2}\pars{-1}^{\ell/2}h^{\ell}    
\end{align}


Then,

$$
\color{#66f}{\large\,{\rm P}_{\ell}\pars{0}}
=\color{#66f}{\large\left\{\begin{array}{lcl}
0 & \mbox{if} & \ell\ \mbox{is odd}
\\[2mm]
\pars{-1}^{\ell/2}{\bracks{\ell - 1}/2 \choose \ell/2}
& \mbox{if} & \ell\ \mbox{is even}
\end{array}\right.}
$$
A: It is possible to prove that
$$P_{{L}} \left( 0 \right) ={\frac { \left( -1 \right) ^{L}
{_2F_1(-L,-L;\,1;\,-1)}}{{2}^{L}}}
$$
Please look the equation (35) at  Legendre Polynomials
It is possible to obtain the simplified form
$$P_{{L}} \left( 0 \right) = \left( -1 \right) ^{L}{-1/2\choose L/2}$$
