Jacksonの6H6和公式

Baileyの${}_6H_6$変換公式
\begin{align} &\H66{1+\frac a2,b,c,d,e,f}{\frac a2,1+a-b,1+a-c,1+a-d,1+a-e,1+a-f}{-1}\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1-d)\Gamma(1-e)\Gamma(1-f)\Gamma(1+2a-d-e-f)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a-b-c)\Gamma(1+a-d-e)\Gamma(1+a-d-f)\Gamma(1+a-e-f)}\\ &\qquad\cdot\H33{b,c,1+2a-d-e-f}{1+a-d,1+a-e,1+a-f}1\\ &\qquad+\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1-b)\Gamma(1-c)\Gamma(1-d)\Gamma(1-e)\Gamma(1-f)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a-b-c)\Gamma(d-a)\Gamma(e-a)\Gamma(f-a)}\\ &\qquad\cdot\frac{\Gamma(2+2a-d-e-f)\Gamma(d+e+f-1-2a)}{\Gamma(2+2a-b-d-e-f)\Gamma(2+2a-c-d-e-f)}\\ &\qquad\cdot\F32{1+a-d-e,1+a-d-f,1+a-e-f}{2+2a-b-d-e-f,2+2a-c-d-e-f}1 \end{align}
において, $b=z-x,c=z-y,d=z+x,e=z+y,f=\frac 12+a-z$とすると,
\begin{align} &\H66{1+\frac a2,\frac 12+a-z,z+x,z-x,z+y,z-y}{\frac a2,\frac 12+z,1+a-z-x,1+a-z+x,1+a-z-y,1+a-z+y}{-1}\\ &=\frac{\Gamma(1+a-z+x)\Gamma(1+a-z+y)\Gamma(1-z-x)\Gamma(1-z-y)\Gamma\left(\frac 12-a+z\right)\Gamma\left(\frac 12+a-x-y-z\right)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a+x+y-2z)\Gamma(1+a-x-y-2z)\Gamma\left(\frac 12-x\right)\Gamma\left(\frac 12-y\right)}\\ &\qquad\cdot\H33{z-x,z-y,\frac 12+a-x-y-z}{1+a-z-x,1+a-z-y,\frac 12+z}1\\ &\qquad+\frac{\Gamma(1+a-z+x)\Gamma(1+a-z+y)\Gamma(1-z+x)\Gamma(1-z+y)\Gamma(1-z-x)\Gamma(1-z-y)\Gamma\left(\frac 12-a+z\right)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a+x+y-2z)\Gamma(z+x-a)\Gamma(z+y-a)\Gamma\left(\frac 12-z\right)}\\ &\qquad\cdot\frac{\Gamma(\frac 32+a-x-y-z)\Gamma(x+y+z-a-\frac 12)}{\Gamma\left(\frac 32+a-2z-x\right)\Gamma\left(\frac 32+2a-2z-y\right)}\F32{1+a-x-y-2z,\frac 12-x,\frac 12-y}{\frac 32+a-2z-x,\frac32+a-2z-y}1 \end{align}
ここで, Dixonの和公式の両側類似
\begin{align} &\H33{b,c,d}{1+a-b,1+a-c,1+a-d}1\\ &=\frac{\Gamma(1-b)\Gamma(1-c)\Gamma(1-d)\Gamma\left(1-\frac a2\right)\Gamma\left(1+\frac a2\right)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)}\\ &\qquad\cdot\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma\left(1+\frac{3a}2-b-c-d\right)}{\Gamma\left(1+\frac a2-b\right)\Gamma\left(1+\frac a2-c\right)\Gamma\left(1+\frac a2-d\right)} \end{align}
より,
\begin{align} &\H33{z-x,z-y,\frac 12+a-x-y-z}{1+a-z-x,1+a-z-y,\frac 12+z}1\\ &=\frac{\Gamma(1-z+x)\Gamma(1-z+y)\Gamma\left(\frac 12-a+x+y+z\right)\Gamma\left(1-\frac {a-x-y}2\right)\Gamma\left(1+\frac{a-x-y}2\right)}{\Gamma(1+a-x-y)\Gamma(1-a+x+y)\Gamma(1+a-2z)\Gamma\left(\frac 12+x\right)\Gamma\left(\frac 12+y\right)}\\ &\qquad\cdot\frac{\Gamma(1+a-z-x)\Gamma(1+a-z-y)\Gamma\left(\frac 12+z\right)\Gamma\left(\frac{1+a+x+y}2-z\right)}{\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(\frac{1-a+x+y}2+z\right)}\\ &=\frac{\Gamma(1-z+x)\Gamma(1-z+y)\Gamma\left(\frac 12-a+x+y+z\right)\Gamma(1+a-z-x)\Gamma(1+a-z-y)\Gamma\left(\frac 12+z\right)\Gamma\left(\frac{1+a+x+y}2-z\right)}{\Gamma(1+a-2z)\Gamma\left(\frac 12+x\right)\Gamma\left(\frac 12+y\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(\frac{1-a+x+y}2+z\right)}\cos\frac{\pi(a-x-y)}2 \end{align}
である. また, Dixonの和公式 より
\begin{align} &\F32{1+a-x-y-2z,\frac 12-x,\frac 12-y}{\frac 32+a-2z-x,\frac32+a-2z-y}1\\ &=\frac{\Gamma\left(\frac 32+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(\frac 32+a-2z-x\right)\Gamma\left(\frac 32+a-2z-y\right)\Gamma\left(\frac 12+\frac a2-z+\frac{x+y}2\right)}{\Gamma(2+a-x-y-2z)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma(1+a-2z)} \end{align}
であるから, これらを代入すると,
\begin{align} &\H66{1+\frac a2,\frac 12+a-z,z+x,z-x,z+y,z-y}{\frac a2,\frac 12+z,1+a-z-x,1+a-z+x,1+a-z-y,1+a-z+y}{-1}\\ &=\frac{\Gamma(1+a-z+x)\Gamma(1+a-z+y)\Gamma(1-z-x)\Gamma(1-z-y)\Gamma\left(\frac 12-a+z\right)\Gamma\left(\frac 12+a-x-y-z\right)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a+x+y-2z)\Gamma(1+a-x-y-2z)\Gamma\left(\frac 12-x\right)\Gamma\left(\frac 12-y\right)}\\ &\qquad\cdot\frac{\Gamma(1-z+x)\Gamma(1-z+y)\Gamma\left(\frac 12-a+x+y+z\right)\Gamma(1+a-z-x)\Gamma(1+a-z-y)\Gamma\left(\frac 12+z\right)\Gamma\left(\frac{1+a+x+y}2-z\right)}{\Gamma(1+a-2z)\Gamma\left(\frac 12+x\right)\Gamma\left(\frac 12+y\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(\frac{1-a+x+y}2+z\right)}\cos\frac{\pi(a-x-y)}2\\ &\qquad+\frac{\Gamma(1+a-z+x)\Gamma(1+a-z+y)\Gamma(1-z+x)\Gamma(1-z+y)\Gamma(1-z-x)\Gamma(1-z-y)\Gamma\left(\frac 12-a+z\right)}{\Gamma(1+a)\Gamma(1-a)\Gamma(1+a+x+y-2z)\Gamma(z+x-a)\Gamma(z+y-a)\Gamma\left(\frac 12-z\right)}\\ &\qquad\cdot\frac{\Gamma(\frac 32+a-x-y-z)\Gamma(x+y+z-a-\frac 12)}{\Gamma\left(\frac 32+a-2z-x\right)\Gamma\left(\frac 32+2a-2z-y\right)}\frac{\Gamma\left(\frac 32+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(\frac 32+a-2z-x\right)\Gamma\left(\frac 32+a-2z-y\right)\Gamma\left(\frac 12+\frac a2-z+\frac{x+y}2\right)}{\Gamma(2+a-x-y-2z)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma(1+a-2z)}\\ &=\frac{\Gamma(1+a-z+x)\Gamma(1+a-z+y)\Gamma(1-z-x)\Gamma(1-z-y)\Gamma\left(\frac 12-a+z\right)\Gamma\left(\frac 12+a-x-y-z\right)}{\pi^2\Gamma(1+a)\Gamma(1-a)\Gamma(1+a+x+y-2z)\Gamma(1+a-x-y-2z)}\cos{\pi x}\cos{\pi y}\\ &\qquad\cdot\frac{\Gamma(1-z+x)\Gamma(1-z+y)\Gamma\left(\frac 12-a+x+y+z\right)\Gamma(1+a-z-x)\Gamma(1+a-z-y)\Gamma\left(\frac 12+z\right)\Gamma\left(\frac{1+a+x+y}2-z\right)}{\Gamma(1+a-2z)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(\frac{1-a+x+y}2+z\right)}\cos\frac{\pi(a-x-y)}2\\ &\qquad+\frac{\Gamma(1+a-z+x)\Gamma(1+a-z+y)\Gamma(1+a-z-x)\Gamma(1+a-z-y)\Gamma(1-z+x)\Gamma(1-z+y)\Gamma(1-z-x)\Gamma(1-z-y)\Gamma\left(\frac 12-a+z\right)\Gamma\left(\frac 12+z\right)}{\pi^3\Gamma(1+a)\Gamma(1-a)\Gamma(1+a+x+y-2z)}\\ &\qquad\cdot\frac{\Gamma(\frac 32+a-x-y-z)\Gamma(x+y+z-a-\frac 12)\Gamma\left(\frac 32+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(\frac 12+\frac a2-z+\frac{x+y}2\right)}{\Gamma(2+a-x-y-2z)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma(1+a-2z)}\\ &\qquad\cdot\cos\pi z\sin\pi(z+x-a)\sin\pi(z+y-a)\\ &=\frac{A}{\Gamma(1+a+x+y-2z)\Gamma(1+a-x-y-2z)}\frac{\cos{\pi x}\cos{\pi y}}{\cos\pi(a-x-y-z)}\\ &\qquad\cdot\frac{\Gamma\left(\frac{1+a+x+y}2-z\right)}{\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(\frac{1-a+x+y}2+z\right)}\cos\frac{\pi(a-x-y)}2\\ &\qquad-\frac{A\Gamma\left(\frac 32+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(\frac 12+\frac a2-z+\frac{x+y}2\right)}{\pi\Gamma(1+a+x+y-2z)\Gamma(2+a-x-y-2z)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)}\\ &\qquad\cdot\frac{\cos\pi z\sin\pi(z+x-a)\sin\pi(z+y-a)}{\cos\pi(a-x-y-z)}\\ &=\frac{A\pi}{2^{2a-4z}\Gamma\left(\frac{1+a-x-y}2-z\right)\Gamma\left(1+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)}\frac{\cos{\pi x}\cos{\pi y}}{\cos\pi(a-x-y-z)}\\ &\qquad\cdot\frac{1}{\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(\frac{1-a+x+y}2+z\right)}\cos\frac{\pi(a-x-y)}2\\ &\qquad-\frac{A}{2^{2a+1-4z}\Gamma\left(1+\frac a2-z+\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)}\\ &\qquad\cdot\frac{\cos\pi z\sin\pi(z+x-a)\sin\pi(z+y-a)}{\cos\pi(a-x-y-z)}\\ &=\frac{A}{2^{2a+1-4z}\Gamma\left(1+\frac a2-z+\frac x2+\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z+\frac x2-\frac y2\right)\Gamma\left(1+\frac a2-z-\frac x2+\frac y2\right)\cos\pi(a-x-y-z)}\frac{}{}\\ &\qquad\cdot\left(2\cos{\pi x}\cos{\pi y}\cos\frac{\pi(a-x-y)}2\cos\pi\left(z-\frac{a-x-y}2\right)-\cos\pi z\sin\pi(z+x-a)\sin\pi(z+y-a)\right) \end{align}
ここで,
\begin{align} A&:=\frac{\Gamma(1-z-x)\Gamma(1-z+x)\Gamma(1-z-y)\Gamma(1-z+y)\Gamma\left(\frac 12+z\right)\Gamma\left(\frac 12+z-a\right)}{\pi\Gamma(1+a)\Gamma(1-a)\Gamma(1+a-2z)}\\ &\qquad\cdot\Gamma(1+a-z-x)\Gamma(1+a-z+x)\Gamma(1+a-z-y)\Gamma(1+a-z+y) \end{align}
とした. 積和の公式や加法定理を用いて変形すると,
\begin{align} &2\cos{\pi x}\cos{\pi y}\cos\frac{\pi(a-x-y)}2\cos\pi\left(z-\frac{a-x-y}2\right)-\cos\pi z\sin\pi(z+x-a)\sin\pi(z+y-a)\\ &=\cos{\pi x}\cos{\pi y}(\cos\pi z+\cos\pi(a-x-y-z))-\cos\pi z\sin\pi(z+x-a)\sin\pi(z+y-a)\\ &=\cos{\pi x}\cos{\pi y}\cos\pi(a-x-y-z)+\cos\pi z(\cos\pi x\cos\pi y-\sin\pi(z+x-a)\sin\pi(z+y-a))\\ &=\cos{\pi x}\cos{\pi y}\cos\pi(a-x-y-z)+\frac 12\cos\pi z(\cos\pi(x+y)+\cos\pi(x-y)+\cos\pi(x+y+2z-2a)-\cos\pi(x-y))\\ &=\cos{\pi x}\cos{\pi y}\cos\pi(a-x-y-z)+\frac 12\cos\pi z(\cos\pi(x+y)+\cos\pi(x+y+2z-2a))\\ &=\cos{\pi x}\cos{\pi y}\cos\pi(a-x-y-z)+\cos\pi z\cos\pi(x+y+z-a)\cos\pi(z-a) \end{align}
であるから, これを代入して定理を得る.