SingalによるWatsonの3F2和公式の二重類似

Srivastavaによる展開公式 より,
\begin{align} &\sum_{0\leq m,n}\frac{(a,2b,2c)_m(a,2b',2c')_n}{(b+c+\frac 12)m!(b'+c'+\frac 12)_nn!(2a)_{m+n}}\\ &=\sum_{0\leq n}\frac{(a,2b,2c,a',2b',2c')_n}{(2a+n-1)_n(2a)_{2n}\left(b+c+\frac 12,b'+c'+\frac 12\right)_n}\frac{(-1)^n}{n!}\\ &\cdot\F32{a+n,2b+n,2c+n}{b+c+\frac 12+n,2a+2n}{1}\F32{a+n,2b'+n,2c'+n}{b'+c'+\frac 12+n,2a+2n}{1} \end{align}
である. Watsonの${}_3F_2$和公式 より,
\begin{align} \F32{a+n,2b+n,2c+n}{b+c+\frac 12+n,2a+2n}{1}&=\frac{\Gamma\left(\frac 12\right)\Gamma\left(b+c+\frac 12+n\right)\Gamma\left(a+n+\frac 12\right)\Gamma\left(a+\frac 12-b-c\right)}{\Gamma\left(b+\frac{n+1}2\right)\Gamma\left(c+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b\right)\Gamma\left(a+\frac {n+1}2-c\right)}\\ \F32{a+n,2b'+n,2c'+n}{b'+c'+\frac 12+n,2a+2n}{1}&=\frac{\Gamma\left(\frac 12\right)\Gamma\left(b'+c'+\frac 12+n\right)\Gamma\left(a+n+\frac 12\right)\Gamma\left(a+\frac 12-b'-c'\right)}{\Gamma\left(b'+\frac{n+1}2\right)\Gamma\left(c'+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b'\right)\Gamma\left(a+\frac {n+1}2-c'\right)} \end{align}
を用いると,
\begin{align} &\sum_{0\leq n}\frac{(a,2b,2c,a,2b',2c')_n}{(2a+n-1)_n(2a)_{2n}\left(b+c+\frac 12,b'+c'+\frac 12\right)_n}\frac{(-1)^n}{n!}\\ &\cdot\F32{a+n,2b+n,2c+n}{b+c+\frac 12+n,2a+2n}{1}\F32{a+n,2b'+n,2c'+n}{b'+c'+\frac 12+n,2a+2n}{1}\\ &=\sum_{0\leq n}\frac{(a,2b,2c,a,2b',2c')_n}{(2a+n-1)_n(2a)_{2n}\left(b+c+\frac 12,b'+c'+\frac 12\right)_n}\frac{(-1)^n}{n!}\\ &\cdot\frac{\Gamma\left(\frac 12\right)\Gamma\left(b+c+\frac 12+n\right)\Gamma\left(a+n+\frac 12\right)\Gamma\left(a+\frac 12-b-c\right)}{\Gamma\left(b+\frac{n+1}2\right)\Gamma\left(c+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b\right)\Gamma\left(a+\frac {n+1}2-c\right)}\\ &\cdot\frac{\Gamma\left(\frac 12\right)\Gamma\left(b'+c'+\frac 12+n\right)\Gamma\left(a+n+\frac 12\right)\Gamma\left(a+\frac 12-b'-c'\right)}{\Gamma\left(b'+\frac{n+1}2\right)\Gamma\left(c'+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b'\right)\Gamma\left(a+\frac {n+1}2-c'\right)}\\ &=\pi\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(a+\frac 12-b-c\right)\Gamma\left(a+\frac 12-b'-c'\right)\Gamma\left(a+\frac 12\right)^2\\ &\cdot\sum_{0\leq n}\frac{2a+2n-1}{2a-1}\frac{(2a-1,2b,2c,2b',2c')_n}{2^{4n}}\frac{(-1)^n}{n!}\\ &\cdot\frac{1}{\Gamma\left(b+\frac{n+1}2\right)\Gamma\left(c+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b\right)\Gamma\left(a+\frac {n+1}2-c\right)}\\ &\cdot\frac{1}{\Gamma\left(b'+\frac{n+1}2\right)\Gamma\left(c'+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b'\right)\Gamma\left(a+\frac {n+1}2-c'\right)}\\ \end{align}
ここで,
\begin{align} &\sum_{0\leq n}\frac{2a+2n-1}{2a-1}\frac{(2a-1,2b,2c,2b',2c')_n}{2^{4n}}\frac{(-1)^n}{n!}\\ &\cdot\frac{1}{\Gamma\left(b+\frac{n+1}2\right)\Gamma\left(c+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b\right)\Gamma\left(a+\frac {n+1}2-c\right)}\\ &\cdot\frac{1}{\Gamma\left(b'+\frac{n+1}2\right)\Gamma\left(c'+\frac {n+1}2\right)\Gamma\left(a+\frac {n+1}2-b'\right)\Gamma\left(a+\frac {n+1}2-c'\right)}\\ &=\sum_{0\leq n}\frac{2a+4n-1}{2a-1}\frac{(2a-1,2b,2c,2b',2c')_{2n}}{2^{8n}}\frac{1}{(2n)!}\\ &\cdot\frac{1}{\Gamma\left(b+\frac{1}2+n\right)\Gamma\left(c+\frac {1}2+n\right)\Gamma\left(a+\frac {1}2-b+n\right)\Gamma\left(a+\frac {1}2-c+n\right)}\\ &\cdot\frac{1}{\Gamma\left(b'+\frac{1}2+n\right)\Gamma\left(c'+\frac {1}2+n\right)\Gamma\left(a+\frac {1}2-b'+n\right)\Gamma\left(a+\frac {1}2-c'+n\right)}\\ &-\sum_{0\leq n}\frac{2a+4n+1}{2a-1}\frac{(2a-1,2b,2c,2b',2c')_{2n+1}}{2^{8n+4}}\frac{1}{(2n+1)!}\\ &\cdot\frac{1}{\Gamma\left(b+n+1\right)\Gamma\left(c+n+1\right)\Gamma\left(a+1-b+n\right)\Gamma\left(a+1-c+n\right)}\\ &\cdot\frac{1}{\Gamma\left(b'+n+1\right)\Gamma\left(c'+n+1\right)\Gamma\left(a+1-b'+n\right)\Gamma\left(a+1-c'+n\right)}\\ &=\frac 1{\Gamma\left(b+\frac 12\right)\Gamma\left(c+\frac 12\right)\Gamma\left(a+\frac 12-b\right)\Gamma\left(a+\frac 12-c\right)\Gamma\left(b'+\frac 12\right)\Gamma\left(c'+\frac 12\right)\Gamma\left(a+\frac 12-b'\right)\Gamma\left(a+\frac 12-c'\right)}\\ &\cdot\sum_{0\leq n}\frac{2a+4n-1}{2a-1}\frac{\left(a-\frac 12,a,b,c,b',c'\right)_n}{\left(a+\frac 12-b,a+\frac 12-c,a+\frac 12-b',a+\frac 12-c',\frac 12\right)_nn!}\\ &-\frac 1{\Gamma(b)\Gamma(c)\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(b')\Gamma(c')\Gamma(1+a-b')\Gamma(1+a-c')}\\ &\cdot\sum_{0\leq n}(2a+4n+1)\frac{\left(a,a+\frac 12,b+\frac 12,c+\frac 12,b'+\frac 12,c'+\frac 12\right)_n}{\left(1+a-b,1+a-c,1+a-b',1+a-c',\frac 32\right)_nn!} \end{align}
よって, これを代入して定理を得る.

定理1とDougallの${}_5F_4$和公式より,
\begin{align} &\sum_{0\leq k,l}\frac{(a,2b,2c)_k}{k!\left(b+c+\frac 12\right)_k}\frac{(a,2b',2c')_l}{l!\left(b'+c'+\frac 12\right)_l}\frac 1{(2a)_{k+l}}\\ &=\frac{\pi\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(a+\frac 12-b-c\right)\Gamma\left(a+\frac 12-b'-c'\right)\Gamma\left(a+\frac 12\right)^2}{\Gamma\left(b+\frac 12\right)\Gamma\left(c+\frac 12\right)\Gamma\left(a+\frac 12-b\right)\Gamma\left(a+\frac 12-c\right)\Gamma\left(b'+\frac 12\right)\Gamma\left(c'+\frac 12\right)\Gamma\left(a+\frac 12-b'\right)\Gamma\left(a+\frac 12-c'\right)}\\ &\cdot\F54{a-\frac 12,\frac a2+\frac 34,a,c,c'}{\frac a2-\frac 14,\frac 12,\frac 12+a-c,\frac 12+a-c'}1\\ &-\frac {2\pi\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(a+\frac 12-b-c\right)\Gamma\left(a+\frac 12-b'-c'\right)\Gamma\left(a+\frac 12\right)\Gamma\left(a+\frac 32\right)}{\Gamma(b)\Gamma(c)\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(b')\Gamma(c')\Gamma(1+a-b')\Gamma(1+a-c')}\\ &\cdot\F54{a+\frac 12,\frac a2+\frac 54,a,c+\frac 12,c'+\frac 12}{\frac a2+\frac 14,\frac 32,1+a-c,1+a-c'}1\\ &=\frac{\pi\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(b'-c\right)\Gamma\left(b-c'\right)\Gamma\left(a+\frac 12\right)^2}{\Gamma\left(b\right)\Gamma\left(b+\frac 12\right)\Gamma\left(b'\right)\Gamma\left(b'+\frac 12\right)\Gamma\left(c+\frac 12\right)\Gamma\left(a+\frac 12-c\right)\Gamma\left(c'+\frac 12\right)\Gamma\left(a+\frac 12-c'\right)}\\ &\cdot\frac{\Gamma\left(\frac 12\right)\Gamma\left(\frac 12+a-c\right)\Gamma\left(\frac 12+a-c'\right)\Gamma\left(\frac 12-c-c'\right)}{\Gamma\left(\frac 12+a\right)\Gamma\left(\frac 12-c\right)\Gamma\left(\frac 12-c'\right)\Gamma\left(\frac 12+a-c-c'\right)}\\ &-\frac {2\pi\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(b'-c\right)\Gamma\left(b-c'\right)\Gamma\left(a+\frac 12\right)\Gamma\left(a+\frac 32\right)}{\Gamma(b)\Gamma\left(b+\frac 12\right)\Gamma(b')\Gamma\left(b'+\frac 12\right)\Gamma(c)\Gamma(1+a-c)\Gamma(c')\Gamma(1+a-c')}\\ &\cdot\frac{\Gamma\left(\frac 32\right)\Gamma(1+a-c)\Gamma(1+a-c')\Gamma\left(\frac 12-c-c'\right)}{\Gamma\left(\frac 32+a\right)\Gamma(1-c)\Gamma(1-c')\Gamma\left(\frac 12+a-c-c'\right)}\\ &=\frac{\pi^{\frac 32}\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(b'-c\right)\Gamma\left(b-c'\right)\Gamma\left(a+\frac 12\right)\Gamma\left(\frac 12-c-c'\right)}{\Gamma\left(b\right)\Gamma\left(b+\frac 12\right)\Gamma\left(b'\right)\Gamma\left(b'+\frac 12\right)\Gamma\left(c+\frac 12\right)\Gamma\left(c'+\frac 12\right)\Gamma\left(\frac 12-c\right)\Gamma\left(\frac 12-c'\right)\Gamma\left(\frac 12+a-c-c'\right)}\\ &-\frac {\pi^{\frac 32}\Gamma\left(b+c+\frac 12\right)\Gamma\left(b'+c'+\frac 12\right)\Gamma\left(b'-c\right)\Gamma\left(b-c'\right)\Gamma\left(a+\frac 12\right)\Gamma\left(\frac 12-c-c'\right)}{\Gamma(b)\Gamma\left(b+\frac 12\right)\Gamma(b')\Gamma\left(b'+\frac 12\right)\Gamma(c)\Gamma(c')\Gamma(1-c)\Gamma(1-c')\Gamma\left(\frac 12+a-c-c'\right)} \end{align}