Holdemanの展開公式

Legendre多項式の直交性より,
$$\begin{array}{rl}{}_2{F}_1[\begin{array}{c}a,b\\ 1\end{array};\frac{1+x}{2}]& =\sum _{0\le n}{a}_n{P}_n(x)\end{array}$$
と展開したときの係数は
$$\begin{array}{rl}{a}_n& =\frac{2n+1}{2}{\int }_{-1}^1{}_2{F}_1[\begin{array}{c}a,b\\ 1\end{array};\frac{1+x}{2}]P_n(x)dx\end{array}$$
で与えられる. ここで, Rodriguesの公式
$$\begin{array}{rl}{P}_n(x)& =\frac{1}{2^{n}n!}{\left(\frac{d}{dx}\right)}^n(x^2-1{)}^n\end{array}$$
を用いて, $n$回部分積分してベータ積分により,
$$\begin{array}{rl}{a}_n& =\frac{1}{2^{n}n!}\frac{2n+1}{2}{\int }_{-1}^1{}_2{F}_1[\begin{array}{c}a,b\\ 1\end{array};\frac{1+x}{2}]{\left(\frac{d}{dx}\right)}^n(x^2-1{)}^{n}dx\\ & =\frac{(-1{)}^n}{2^{n}n!}\frac{2n+1}{2}{\int }_{-1}^1{\left(\frac{d}{dx}\right)}^n\left({}_2{F}_1[\begin{array}{c}a,b\\ 1\end{array};\frac{1+x}{2}]\right)(x^2-1{)}^{n}dx\\ & =\frac{(-1{)}^n(a,b{)}_n}{2^{2n}n{!}^2}\frac{2n+1}{2}{\int }_{-1}^1{}_2{F}_1[\begin{array}{c}a+n,b+n\\ 1+n\end{array};\frac{1+x}{2}](x^2-1{)}^{n}dx\\ & =\frac{(-1{)}^n(a,b{)}_n}{2^{2n}n{!}^2}\frac{2n+1}{2}\sum _{0\le k}\frac{(a+n,b+n{)}_k}{k!(1+n{)}_k}\frac{1}{2^k}{\int }_{-1}^1(x-1{)}^n(x+1{)}^{n+k}dx\\ & =\frac{(a,b{)}_n}{2^{2n}n{!}^2}\frac{2n+1}{2}\frac{2^{2n+1}n{!}^2}{(2n+1)!}\sum _{0\le k}\frac{(a+n,b+n{)}_k}{k!(2n+2{)}_k}\\ & =\frac{(a,b{)}_n}{(2n)!}\sum _{0\le k}\frac{(a+n,b+n{)}_k}{k!(2n+2{)}_k}\end{array}$$
ここで, Gaussの超幾何定理より,
$$\begin{array}{rl}& \frac{(a,b{)}_n}{(2n)!}\sum _{0\le k}\frac{(a+n,b+n{)}_k}{k!(2n+2{)}_k}\\ & =\frac{(2n+1)!\mathrm{\Gamma }(2-a-b)}{\mathrm{\Gamma }(2-a+n)\mathrm{\Gamma }(2-b+n)}\end{array}$$
だから,
$$\begin{array}{rl}{a}_n& =\frac{\mathrm{\Gamma }(2-a-b)}{\mathrm{\Gamma }(2-a)\mathrm{\Gamma }(2-b)}\frac{(2n+1)(a,b{)}_n}{(2-a,2-b{)}_n}\end{array}$$
となって定理を得る.