Integral including Legendre and radical

Question

I would appreciate if you help me with the integral $\int\sqrt{\beta^2 - x^2} P_n(x) dx$. As simple as it may look, I could not find it in the table of integrals of different handbooks.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int\root{\beta^{2} - x^{2}}{\rm P}_n\pars{x}\,\dd x:\ {\large ?}}$

\begin{align} &{1 \over \root{1 - 2xh + h^{2}}}=\sum_{n = 0}^{\infty}h^{n}{\rm P}_{n}\pars{x} \\[3mm]&\imp\quad \int{\root{\beta^{2} - x^{2}} \over \root{1 - 2xh + h^{2}}}\,\dd x =\sum_{n = 0}^{\infty}h^{n}\int\root{\beta^{2} - x^{2}}{\rm P}_{n}\pars{x}\,\dd x \end{align} Integrates the left hand side and expands in power of $\ds{h}$. Mathematica can do that !!!.