Explicit Formula for the Integral of the Legendre Polynomial

Question

Question: Expand the sign function between $-1$ and $1$ in a series of Legendre polynomials. Obtain an explicit formula for the expansion coefficients.

Attempt: I found that

$$a_l=(2l+1)\int_0^1 P_l(x)$$

where $a_l$ is the coefficient of degree l. However, I am still confused how to find an explicit formula for this value. Can someone please help me?

Answer 1

Hint: use the recurrence relation:

$$(2 \ell+1) P_{\ell}(x) = \frac{d}{dx} [P_{\ell+1}(x) - P_{\ell-1}(x)]$$

Answer 2

Note that,

$$ a_0=1,\, a_{2 l}=0 .$$

For odd $l$, one can have the formula

$$ a_{2l-1}= {\frac { \left( -1 \right)^{l-1} \left( 4\,l-1 \right) \Gamma \left( l-\frac{1}{2} \right) }{2\sqrt {\pi }\,\Gamma \left( l+1 \right) }} \quad l \in \mathbb{N}. $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\on{P}_{\ell}:\ Order\mbox{-}\ell\ Polynomial}$.

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\begin{align} a_{\ell} & \equiv \bbox[5px,#ffd]{\left.\pars{2\ell + 1} \int_{0}^{1}\on{P}_{\ell}\pars{x}\,\dd x \,\right\vert_{\,\ell\ \in\ \mathbb{N}_{\,\geq\ 0}}} \\[5mm] & = \pars{2\ell + 1} \int_{0}^{1}\braces{\bracks{h^{\ell}}{1 \over \root{1 - 2xh + h^{2}}}}\,\dd x \\[5mm] & = \pars{2\ell + 1}\bracks{h^{\ell}} \int_{0}^{1}{\dd x \over\root{1 - 2xh + h^{2}}} \\[5mm] = &\ \pars{2\ell + 1} \bracks{h^{\ell}}{\root{1 + h^{2}} + h - 1 \over h} \\[5mm] & = \pars{2\ell + 1}\bracks{\ell\ odd} \bracks{h^{\ell + 1}}\root{1 + h^{2}} + \delta_{\ell 0} \\[5mm] & = \delta_{\ell 0} + \pars{2\ell + 1}\bracks{\ell\ odd} {1/2 \choose \bracks{\ell + 1}/2} \end{align}