超幾何関数の積2

$$\begin{array}{rl}& {}_1{F}_1[\begin{array}{c}a\\ b\end{array};x]{}_1{F}_1[\begin{array}{c}1+a-b\\ 2-b\end{array};-x]\\ & =\sum _{0\le n}{x}^n\sum _{k=0}^n\frac{(a{)}_k}{k!(b{)}_k}\frac{(1+a-b{)}_{n-k}}{(n-k)!(2-b{)}_{n-k}}(-1{)}^{n-k}\\ & =\sum _{0\le n}\frac{(1+a-b{)}_n}{n!(2-b{)}_n}(-x{)}^n{}_3{F}_2[\begin{array}{c}b-n-1,a,-n\\ b-a-n,b\end{array};1]\end{array}$$
ここで, Dixonの和公式より,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}a,b,-n\\ 1+a-b,1+a+n\end{array};1]& =\frac{{(1+a,1+\frac{a}{2}-b)}_n}{{(1+\frac{a}{2},1+a-b)}_n}\end{array}$$
であるから, $a\mapsto b-n-1,b\mapsto a$として,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}b-n-1,a,-n\\ b-a-n,b\end{array};1]& =\frac{{(b-n,\frac{b-n+1}{2}-a)}_n}{{(\frac{b-n+1}{2},b-a-n)}_n}\\ & =\frac{{(1-b,\frac{b-n+1}{2}-a)}_n}{{(\frac{b-n+1}{2},1+a-b)}_n}\end{array}$$
$n=2k$のとき,
$$\begin{array}{rl}\frac{{(1-b,\frac{b-n+1}{2}-a)}_n}{n!{(\frac{b-n+1}{2},2-b)}_n}& =\frac{{(1-b,\frac{b+1}{2}-a-k)}_{2k}}{{(1,\frac{b+1}{2}-k,2-b)}_{2k}}\\ & =\frac{{(\frac{b+1}{2}-a,\frac{1-b}{2}+a)}_k}{4^k{(1,\frac{1}{2},\frac{b+1}{2},\frac{3-b}{2})}_k}\end{array}$$
$n=2k+1$のとき,
$$\begin{array}{rl}\frac{{(1-b,\frac{b-n+1}{2}-a)}_n}{n!{(\frac{b-n+1}{2},2-b)}_n}& =\frac{{(1-b,\frac{b}{2}-a-k)}_{2k+1}}{{(1,\frac{b}{2}-k,2-b)}_{2k+1}}\\ & =\frac{(1-b)(b-2a)}{b(2-b)}\frac{(2-b{)}_{2k}{(\frac{b}{2}-a-k,\frac{b}{2}-a+1)}_k}{{(2,3-b)}_{2k}{(\frac{b}{2}-k,\frac{b}{2}+1)}_k}\\ & =\frac{(1-b)(b-2a)}{b(2-b)}\frac{{(1+a-\frac{b}{2},\frac{b}{2}-a+1)}_k}{4^k{(1,\frac{3}{2},2-\frac{b}{2},\frac{b}{2}+1)}_k}\end{array}$$
となるから, 定理を得る.

左辺は奇数次の項がないので,
$$\begin{array}{rl}{}_2{F}_0[\begin{array}{c}a,b\\ -\end{array};x]{}_2{F}_0[\begin{array}{c}a,b\\ -\end{array};-x]& =\sum _{0\le n}{x}^{2n}\sum _{k=0}^{2n}\frac{(a,b{)}_k}{k!}\frac{(a,b{)}_{2n-k}}{(2n-k)!}(-1{)}^k\\ & =\sum _{0\le n}\frac{(a,b{)}_{2n}}{(2n)!}{x}^{2n}{}_3{F}_2[\begin{array}{c}-2n,a,b\\ 1-2n-a,1-2n-b\end{array};1]\end{array}$$
ここで, Dixonの恒等式より,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}-2n,a,b\\ 1-2n-a,1-2n-b\end{array};1]& =(-1{)}^n\frac{\mathrm{\Gamma }(1-2n-a)\mathrm{\Gamma }(1-2n-b)\mathrm{\Gamma }(1-n-a-b)}{\mathrm{\Gamma }(1-n-a)\mathrm{\Gamma }(1-n-b)\mathrm{\Gamma }(1-2n-a-b)}\frac{(2n)!}{n!}\\ & =\frac{(a,b{)}_n(a+b,1{)}_{2n}}{(a,b{)}_{2n}(a+b,1{)}_n}\end{array}$$
であるから, これを代入して定理を得る.

$$\begin{array}{rl}{}_2{F}_0[\begin{array}{c}a,1-a\\ -\end{array};x]{}_2{F}_0[\begin{array}{c}b,1-b\\ -\end{array};-x]& =\sum _{0\le n}(-x{)}^n\sum _{k=0}^n\frac{(a,1-a{)}_k}{k!}\frac{(b,1-b{)}_{n-k}}{(n-k)!}(-1{)}^k\\ & =\sum _{0\le n}\frac{(b,1-b{)}_n}{n!}(-x{)}^n{}_3{F}_2[\begin{array}{c}a,1-a,-n\\ 1-n-b,b-n\end{array};1]\end{array}$$
ここで, Whippleの和公式 より,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}a,1-a,-n\\ 1-n-b,b-n\end{array};1]& =2^{2n+1}\pi \frac{\mathrm{\Gamma }(1-n-b)\mathrm{\Gamma }(b-n)}{\mathrm{\Gamma }\left(\frac{1-n-b+a}{2}\right)\mathrm{\Gamma }\left(\frac{b-n+a}{2}\right)\mathrm{\Gamma }\left(\frac{2-n-b-a}{2}\right)\mathrm{\Gamma }\left(\frac{1+b-n-a}{2}\right)}\end{array}$$
である. $n=2k$のとき,
$$\begin{array}{rl}& 2^{2n+1}\pi \frac{\mathrm{\Gamma }(1-n-b)\mathrm{\Gamma }(b-n)}{\mathrm{\Gamma }\left(\frac{1-n-b+a}{2}\right)\mathrm{\Gamma }\left(\frac{b-n+a}{2}\right)\mathrm{\Gamma }\left(\frac{2-n-b-a}{2}\right)\mathrm{\Gamma }\left(\frac{1+b-n-a}{2}\right)}\\ & =2^{4k+1}\pi \frac{\mathrm{\Gamma }(1-2k-b)\mathrm{\Gamma }(b-2k)}{\mathrm{\Gamma }(\frac{1-b+a}{2}-k)\mathrm{\Gamma }(\frac{b+a}{2}-k)\mathrm{\Gamma }(\frac{2-b-a}{2}-k)\mathrm{\Gamma }(\frac{1+b-a}{2}-k)}\\ & =2^{4k+1}\pi \frac{\mathrm{\Gamma }(1-b)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }\left(\frac{1-b+a}{2}\right)\mathrm{\Gamma }\left(\frac{b+a}{2}\right)\mathrm{\Gamma }\left(\frac{2-b-a}{2}\right)\mathrm{\Gamma }\left(\frac{1+b-a}{2}\right)}\frac{(1-b,b{)}_{-2k}}{{(\frac{1-b+a}{2},\frac{b+a}{2},\frac{2-b-a}{2},\frac{1+b-a}{2})}_{-k}}\\ & =2^{4k}\frac{{(\frac{1-b+a}{2},\frac{b+a}{2},\frac{2-b-a}{2},\frac{1+b-a}{2})}_k}{(1-b,b{)}_{2k}}\end{array}$$
$n=2k+1$のとき,
$$\begin{array}{rl}& 2^{2n+1}\pi \frac{\mathrm{\Gamma }(1-n-b)\mathrm{\Gamma }(b-n)}{\mathrm{\Gamma }\left(\frac{1-n-b+a}{2}\right)\mathrm{\Gamma }\left(\frac{b-n+a}{2}\right)\mathrm{\Gamma }\left(\frac{2-n-b-a}{2}\right)\mathrm{\Gamma }\left(\frac{1+b-n-a}{2}\right)}\\ & =2^{4k+3}\pi \frac{\mathrm{\Gamma }(-2k-b)\mathrm{\Gamma }(b-2k-1)}{\mathrm{\Gamma }(\frac{-b+a}{2}-k)\mathrm{\Gamma }(\frac{b+a-1}{2}-k)\mathrm{\Gamma }(\frac{1-b-a}{2}-k)\mathrm{\Gamma }(\frac{b-a}{2}-k)}\\ & =(a+b+1)(a-b)2^{4k+1}\pi \frac{\mathrm{\Gamma }(1-b)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }\left(\frac{2-b+a}{2}\right)\mathrm{\Gamma }\left(\frac{b+a+1}{2}\right)\mathrm{\Gamma }\left(\frac{1-b-a}{2}\right)\mathrm{\Gamma }\left(\frac{b-a}{2}\right)}\frac{{(\frac{2-a+b}{2},\frac{3-a-b}{2},\frac{1+a+b}{2},\frac{2+a-b}{2})}_k}{(1-b,b{)}_{2k+1}}\\ & =(a+b+1)(a-b)2^{4k}\frac{{(\frac{2-a+b}{2},\frac{3-a-b}{2},\frac{1+a+b}{2},\frac{2+a-b}{2})}_k}{(1-b,b{)}_{2k+1}}\end{array}$$
となるので, これらを代入して定理を得る.