Whippleの3F2和公式, 6F5和公式

まず, Thomaeの${}_3{F}_2$変換公式
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}a,b,c\\ d,e\end{array};1]& =\frac{\mathrm{\Gamma }(d)\mathrm{\Gamma }(e)\mathrm{\Gamma }(d+e-a-b-c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(d+e-a-b)\mathrm{\Gamma }(d+e-a-c)}{}_3{F}_2[\begin{array}{c}d-a,e-a,d+e-a-b-c\\ d+e-a-b,d+e-a-c\end{array};1]\end{array}$$
より,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}a,1-a,c\\ e,1+2c-e\end{array};1]& =\frac{\mathrm{\Gamma }(e)\mathrm{\Gamma }(1+2c-e)\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(1+c-a)\mathrm{\Gamma }(2c)}{}_3{F}_2[\begin{array}{c}e-a,1+2c-e-a,c\\ 1+c-a,2c\end{array};1]\end{array}$$
ここで, Watsonの${}_3{F}_2$和公式 より,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}e-a,1+2c-e-a,c\\ 1+c-a,2c\end{array};1]& =\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(1+c-a)\mathrm{\Gamma }(c+\frac{1}{2})\mathrm{\Gamma }(a)}{\mathrm{\Gamma }\left(\frac{e-a+1}{2}\right)\mathrm{\Gamma }(1+c-\frac{e+a}{2})\mathrm{\Gamma }(c+\frac{1-e+a}{2})\mathrm{\Gamma }\left(\frac{e+a}{2}\right)}\end{array}$$
であるから, これを代入して, Legendreの倍角公式$\mathrm{\Gamma }(c)\mathrm{\Gamma }(c+\frac{1}{2})=2^{1-2c}\sqrt{\pi }\mathrm{\Gamma }(2c)$を用いて整理すると,
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}a,1-a,c\\ e,1+2c-e\end{array};1]& =\frac{\mathrm{\Gamma }(e)\mathrm{\Gamma }(1+2c-e)\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(1+c-a)\mathrm{\Gamma }(2c)}\cdot \frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(1+c-a)\mathrm{\Gamma }(c+\frac{1}{2})\mathrm{\Gamma }(a)}{\mathrm{\Gamma }\left(\frac{e-a+1}{2}\right)\mathrm{\Gamma }(1+c-\frac{e+a}{2})\mathrm{\Gamma }(c+\frac{1-e+a}{2})\mathrm{\Gamma }\left(\frac{e+a}{2}\right)}\\ & =2^{1-2c}\pi \frac{\mathrm{\Gamma }(e)\mathrm{\Gamma }(1+2c-e)}{\mathrm{\Gamma }\left(\frac{e+a}{2}\right)\mathrm{\Gamma }\left(\frac{1+e-a}{2}\right)\mathrm{\Gamma }(c+\frac{1-e+a}{2})\mathrm{\Gamma }(1+c-\frac{e+a}{2})}\end{array}$$
を得る.

$N$を非負整数として, Whippleの${}_7{F}_6$変換公式
$$\begin{array}{rl}& {}_7{F}_6[\begin{array}{c}a,1+\frac{a}{2},b,c,d,e,-N\\ \frac{a}{2},1+a-b,1+a-c,1+a-d,1+a-e,1+a+N\end{array};1]\\ & =\frac{(1+a,1+a-d-e{)}_N}{(1+a-d,1+a-e{)}_N}{}_4{F}_3[\begin{array}{c}1+a-b-c,d,e,-N\\ 1+a-b,1+a-c,d+e-N-a\end{array};1]\end{array}$$
において$N\to \mathrm{\infty }$とすると,
$$\begin{array}{rl}& {}_6{F}_5[\begin{array}{c}a,1+\frac{a}{2},b,c,d,e\\ \frac{a}{2},1+a-b,1+a-c,1+a-d,1+a-e\end{array};-1]\\ & =\frac{\mathrm{\Gamma }(1+a-d)\mathrm{\Gamma }(1+a-e)}{\mathrm{\Gamma }(1+a)\mathrm{\Gamma }(1+a-d-e)}{}_3{F}_2[\begin{array}{c}1+a-b-c,d,e\\ 1+a-b,1+a-c\end{array};1]\end{array}$$
を得る. よってWhippleの${}_3{F}_2$和公式より,
$$\begin{array}{rl}& {}_6{F}_5[\begin{array}{c}a,1+\frac{a}{2},b,1-b,c,1-c\\ \frac{a}{2},1+a-b,a+b,1+a-c,a+c\end{array};-1]\\ & =\frac{\mathrm{\Gamma }(1+a-c)\mathrm{\Gamma }(a+c)}{\mathrm{\Gamma }(1+a)\mathrm{\Gamma }(a)}{}_3{F}_2[\begin{array}{c}a,c,1-c\\ 1+a-b,a+b\end{array};1]\\ & =\frac{\mathrm{\Gamma }(1+a-c)\mathrm{\Gamma }(a+c)}{\mathrm{\Gamma }(1+a)\mathrm{\Gamma }(a)}{2}^{1-2a}\pi \frac{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(1+a-b)}{\mathrm{\Gamma }\left(\frac{a+b+c}{2}\right)\mathrm{\Gamma }\left(\frac{1+a-b+c}{2}\right)\mathrm{\Gamma }\left(\frac{1+a+b-c}{2}\right)\mathrm{\Gamma }\left(\frac{2+a-b-c}{2}\right)}\\ & =2^{1-2a}\pi \frac{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(1+a-b)\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(1+a-c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(1+a)\mathrm{\Gamma }\left(\frac{a+b+c}{2}\right)\mathrm{\Gamma }\left(\frac{1+a-b+c}{2}\right)\mathrm{\Gamma }\left(\frac{1+a+b-c}{2}\right)\mathrm{\Gamma }\left(\frac{2+a-b-c}{2}\right)}\end{array}$$
となって示される.