ある二重超幾何級数の和公式について3: 5F4型の和

前の記事( ある二重超幾何級数の和公式について2: 2F1型の和 )で${}_2F_1$型の和
\begin{align} \sum_{0\leq n}\frac{(a,b)_n}{n!(c)_n}\F{r+1}r{-n,a_1,\dots,a_r}{b_1,\dots,b_r}{z} \end{align}
のMellin-Barnes型積分表示を示した. 今回はその類似として, Dougallの${}_5F_4$和公式に対応する${}_5F_4$型の和
\begin{align} \sum_{0\leq n}\frac{(2n+a)(a,b,c,d)_n}{n!(1+a-b,1+a-c,1+a-d)_n}\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z} \end{align}
のMellin-Barnes型積分表示を与えたいと思う.

導出

まず,
\begin{align} &\sum_{0\leq n}\frac{(2n+a)(a)_n}{n!}x^n\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z}\\ &=\sum_{0\leq n}\frac{(2n+a)(a)_n}{n!}x^n\sum_{0\leq k}\frac{(-n,a+n,b_1,\dots,b_r)_k}{k!(c_1,\dots,c_{r+1})_k}z^k\\ &=\sum_{0\leq k}\frac{(b_1,\dots,b_r)_k}{k!(c_1,\dots,c_{r+1})_k}(-z)^k\sum_{0\leq n}\frac{(2n+a)(a)_{n+k}}{(n-k)!}x^n\\ &=\sum_{0\leq k}\frac{(b_1,\dots,b_r)_k}{k!(c_1,\dots,c_{r+1})_k}(-zx)^k\sum_{0\leq n}\frac{(2n+2k+a)(a)_{n+2k}}{n!}x^n\qquad n\mapsto n+k\\ &=\sum_{0\leq k}\frac{(a)_{2k}(b_1,\dots,b_r)_k}{k!(c_1,\dots,c_{r+1})_k}(-zx)^k\sum_{0\leq n}\frac{(2n+2k+a)(a+2k)_{n}}{n!}x^n\\ &=\sum_{0\leq k}\frac{(a)_{2k}(b_1,\dots,b_r)_k}{k!(c_1,\dots,c_{r+1})_k}(-zx)^k(2x(a+2k)(1-x)^{-a-2k-1}+(2k+a)(1-x)^{-a-2k})\\ &=(1+x)\sum_{0\leq k}\frac{(a)_{2k+1}(b_1,\dots,b_r)_k}{k!(c_1,\dots,c_{r+1})_k}(-zx)^k(1-x)^{-a-2k-1}\\ &=a(1+x)(1-x)^{-a-1}\F{r+2}{r+1}{\frac{a+1}2,1+\frac a2,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{-\frac{4zx}{(1-x)^2}} \end{align}
となる. よって, 項別積分により,
\begin{align} &\sum_{0\leq n}\frac{(2n+a)(a,b,c,d)_n}{n!(1+a-b,1+a-c,1+a-d)_n}\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z}\\ &=\sum_{0\leq n}\frac{(2n+a)(a)_n}{n!}\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z}\\ &\qquad\cdot\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)}{\Gamma(b)\Gamma(1+a-2b)\Gamma(c)\Gamma(1+a-2c)\Gamma(d)\Gamma(1+a-2d)}\\ &\qquad\cdot\int_{0< u,v,w<1}u^{b+n-1}(1-u)^{a-2b}v^{c+n-1}(1-v)^{a-2c}w^{d+n-1}(1-w)^{a-2d}\,dudvdw\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)}{\Gamma(b)\Gamma(1+a-2b)\Gamma(c)\Gamma(1+a-2c)\Gamma(d)\Gamma(1+a-2d)}\\ &\qquad\cdot\int_{0< u,v,w<1}u^{b-1}(1-u)^{a-2b}v^{c-1}(1-v)^{a-2c}w^{d-1}(1-w)^{a-2d}\\ &\qquad\cdot\sum_{0\leq n}\frac{(2n+a)(a)_n}{n!}(uvw)^n\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z}\,dudvdw\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)}{\Gamma(b)\Gamma(1+a-2b)\Gamma(c)\Gamma(1+a-2c)\Gamma(d)\Gamma(1+a-2d)}\\ &\qquad\cdot\int_{0< u,v,w<1}u^{b-1}(1-u)^{a-2b}v^{c-1}(1-v)^{a-2c}w^{d-1}(1-w)^{a-2d}\\ &\qquad\cdot (1+uvw)(1-uvw)^{-a-1}\F{r+2}{r+1}{\frac{a+1}2,1+\frac a2,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{-\frac{4zuvw}{(1-uvw)^2}}\,dudvdw\\ \end{align}
と書き換えられる. ここで, Mellin-Barnes積分により,
\begin{align} &\F{r+2}{r+1}{\frac{a+1}2,1+\frac a2,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{-\frac{4zuvw}{(1-uvw)^2}}\\ &=\frac{\Gamma(c_1)\cdots\Gamma(c_{r+1})}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(1+\frac a2\right)\Gamma(b_1)\cdots\Gamma(b_r)}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac{a+1}2+s\right)\Gamma\left(1+\frac a2+s\right)\Gamma(b_1+s)\cdots\Gamma(b_r+s)\Gamma(-s)}{\Gamma(c_1+s)\cdots\Gamma(c_{r+1}+s)}\left(\frac{4zuvw}{(1-uvw)^2}\right)^s\,ds \end{align}
と展開してFubiniの定理を用いると,
\begin{align} &\sum_{0\leq n}\frac{(2n+a)(a,b,c,d)_n}{n!(1+a-b,1+a-c,1+a-d)_n}\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z}\\ &=\frac{\Gamma(c_1)\cdots\Gamma(c_{r+1})}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(1+\frac a2\right)\Gamma(b_1)\cdots\Gamma(b_r)}\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)}{\Gamma(b)\Gamma(1+a-2b)\Gamma(c)\Gamma(1+a-2c)\Gamma(d)\Gamma(1+a-2d)}\\ &\qquad\cdot\int_{0< u,v,w<1}u^{b-1}(1-u)^{a-2b}v^{c-1}(1-v)^{a-2c}w^{d-1}(1-w)^{a-2d}\\ &\qquad\cdot (1+uvw)(1-uvw)^{-a-1}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac{a+1}2+s\right)\Gamma\left(1+\frac a2+s\right)\Gamma(b_1+s)\cdots\Gamma(b_r+s)\Gamma(-s)}{\Gamma(c_1+s)\cdots\Gamma(c_{r+1}+s)}\left(\frac{4zuvw}{(1-uvw)^2}\right)^s\,ds\,dudvdw\\ &=\frac{\Gamma(c_1)\cdots\Gamma(c_{r+1})}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(1+\frac a2\right)\Gamma(b_1)\cdots\Gamma(b_r)}\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)}{\Gamma(b)\Gamma(1+a-2b)\Gamma(c)\Gamma(1+a-2c)\Gamma(d)\Gamma(1+a-2d)}\\ &\qquad\cdot\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac{a+1}2+s\right)\Gamma\left(1+\frac a2+s\right)\Gamma(b_1+s)\cdots\Gamma(b_r+s)\Gamma(-s)}{\Gamma(c_1+s)\cdots\Gamma(c_{r+1}+s)}(4z)^s\,ds\\ &\qquad\cdot\int_{0< u,v,w<1}u^{b+s-1}(1-u)^{a-2b}v^{c+s-1}(1-v)^{a-2c}w^{d+s-1}(1-w)^{a-2d}(1+uvw)(1-uvw)^{-a-s-1}\,dudvdw \end{align}
ここで, 項別積分により,
\begin{align} &\int_{0< u,v,w<1}u^{b+s-1}(1-u)^{a-2b}v^{c+s-1}(1-v)^{a-2c}w^{d+s-1}(1-w)^{a-2d}(1+uvw)(1-uvw)^{-a-2s-1}\,dudvdw\\ &=\int_{0< u,v,w<1}u^{b+s-1}(1-u)^{a-2b}v^{c+s-1}(1-v)^{a-2c}w^{d+s-1}(1-w)^{a-2d}\sum_{0\leq n}\frac{(2n+a+2s)(a+2s)_n}{n!}(uvw)^n\,dudvdw\\ &=\sum_{0\leq n}\frac{(2n+a+2s)(a+2s)_n}{n!}\frac{\Gamma(b+s+n)\Gamma(c+s+n)\Gamma(d+s+n)\Gamma(1+a-2b)\Gamma(1+a-2c)\Gamma(1+a-2d)}{\Gamma(1+a-b+s+n)\Gamma(1+a-c+s+n)\Gamma(1+a-d+s+n)}\\ &=a\frac{\Gamma(b+s)\Gamma(c+s)\Gamma(d+s)\Gamma(1+a-2b)\Gamma(1+a-2c)\Gamma(1+a-2d)}{\Gamma(1+a-b+s)\Gamma(1+a-c+s)\Gamma(1+a-d+s)}\F54{a+2s,1+\frac{a+2s}2,b+s,c+s,d+s}{\frac{a+2s}2,1+a-b+s,1+a-c+s,1+a-d+s}1\\ &=a\frac{\Gamma(b+s)\Gamma(c+s)\Gamma(d+s)\Gamma(1+a-2b)\Gamma(1+a-2c)\Gamma(1+a-2d)\Gamma(1+a-b-c-d-s)}{\Gamma(1+a+2s)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)} \end{align}
となる. ここで, 最後の等号はDougallの${}_5F_4$和公式による. これを代入すると,
\begin{align} &\sum_{0\leq n}\frac{(2n+a)(a,b,c,d)_n}{n!(1+a-b,1+a-c,1+a-d)_n}\F{r+2}{r+1}{-n,a+n,b_1,\dots,b_r}{c_1,\dots,c_{r+1}}{z}\\ &=\frac{a\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(c_1)\cdots\Gamma(c_{r+1})}{\Gamma\left(\frac{a+1}2\right)\Gamma\left(1+\frac a2\right)\Gamma(b)\Gamma(c)\Gamma(d)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)\Gamma(b_1)\cdots\Gamma(b_r)}\\ &\qquad\cdot\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac{a+1}2+s\right)\Gamma\left(1+\frac a2+s\right)\Gamma(b+s)\Gamma(c+s)\Gamma(d+s)\Gamma(b_1+s)\cdots\Gamma(b_r+s)\Gamma(-s)\Gamma(1+a-b-c-d-s)}{\Gamma(c_1+s)\cdots\Gamma(c_{r+1}+s)\Gamma(1+a+2s)}(4z)^s\,ds\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(c_1)\cdots\Gamma(c_{r+1})}{\Gamma\left(a\right)\Gamma(b)\Gamma(c)\Gamma(d)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)\Gamma(b_1)\cdots\Gamma(b_r)}\\ &\qquad\cdot\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(b+s)\Gamma(c+s)\Gamma(d+s)\Gamma(b_1+s)\cdots\Gamma(b_r+s)\Gamma(-s)\Gamma(1+a-b-c-d-s)}{\Gamma(c_1+s)\cdots\Gamma(c_{r+1}+s)}z^s\,ds \end{align}
となる(最後の等号はLegendreの倍角公式による). つまり以下が得られた.

定理1において, $z=1, r\mapsto r+1, b_{r+1}=w:=2b+c+d-a-1$として, $c_1,\dots,c_{r+2}$を$1+w-b_1,\dots,1+w-b_r,1+w-c,1+w-d$とすると,
\begin{align} &\sum_{0\leq n}\frac{(2n+a)(a,b,c,d)_n}{n!(1+a-b,1+a-c,1+a-d)_n}\F{r+3}{r+2}{-n,a+n,w,b_1,\dots,b_r}{1+w-c,1+w-d,1+w-b_1,\dots,1+w-b_r}{1}\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(1+w-c)\Gamma(1+w-d)\Gamma(1+w-b_1)\cdots\Gamma(1+w-b_r)}{\Gamma\left(a\right)\Gamma(b)\Gamma(c)\Gamma(d)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)\Gamma(w)\Gamma(b_1)\cdots\Gamma(b_r)}\\ &\qquad\cdot\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(w+s)\Gamma(b+s)\Gamma(c+s)\Gamma(d+s)\Gamma(b_1+s)\cdots\Gamma(b_r+s)\Gamma(-s)\Gamma(b-w-s)}{\Gamma(1+w-c+s)\Gamma(1+w-d+s)\Gamma(1+w-b_1+s)\cdots\Gamma(1+w-b_r+s)}\,ds \end{align}
さらに, $r\mapsto r+1$として, $b_{r+1}=1+\frac w2$とすると, 以下を得る.

Non-terminating Dougallの和公式の書き換え

系1において, $1+2w=b+c+d+e+f$として, $b_1,\dots,b_r$を$e,f$とすると, non-terminating Dougallの和公式のMellin-Barnes積分表示 より,
\begin{align} &\sum_{0\leq n}\frac{(2n+a)(a,b,c,d)_n}{n!(1+a-b,1+a-c,1+a-d)_n}\F65{-n,a+n,w,1+\frac w2,e,f}{\frac w2,1+w-c,1+w-d,1+w-e,1+w-f}{1}\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(1+w-c)\Gamma(1+w-d)\Gamma(1+w-e)\Gamma(1+w-f)}{\Gamma\left(a\right)\Gamma(b)\Gamma(c)\Gamma(d)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)\Gamma(w+1)\Gamma(e)\Gamma(f)}\\ &\qquad\cdot\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{(2s+w)\Gamma(w+s)\Gamma(b+s)\Gamma(c+s)\Gamma(d+s)\Gamma(e+s)\Gamma(f+s)\Gamma(-s)\Gamma(b-w-s)}{\Gamma(1+w-c+s)\Gamma(1+w-d+s)\Gamma(1+w-e+s)\Gamma(1+w-f+s)}\,ds\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(1+w-c)\Gamma(1+w-d)\Gamma(1+w-e)\Gamma(1+w-f)}{\Gamma\left(a\right)\Gamma(b)\Gamma(c)\Gamma(d)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)\Gamma(w+1)\Gamma(e)\Gamma(f)}\\ &\qquad\cdot\frac{\Gamma(b)\Gamma(c)\Gamma(d)\Gamma(e)\Gamma(f)\Gamma(b+c-w)\Gamma(b+d-w)\Gamma(b+e-w)\Gamma(b+f-w)}{\Gamma(1+w-c-d)\Gamma(1+w-c-e)\Gamma(1+w-c-f)\Gamma(1+w-d-e)\Gamma(1+w-d-f)\Gamma(1+w-e-f)}\\ &=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(1+w-c)\Gamma(1+w-d)\Gamma(1+w-e)\Gamma(1+w-f)}{\Gamma\left(a\right)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)\Gamma(w+1)}\\ &\qquad\cdot\frac{\Gamma(b+c-w)\Gamma(b+d-w)\Gamma(b+e-w)\Gamma(b+f-w)}{\Gamma(1+w-c-d)\Gamma(1+w-c-e)\Gamma(1+w-c-f)\Gamma(1+w-d-e)\Gamma(1+w-d-f)\Gamma(1+w-e-f)} \end{align}
となる. つまり以下が得られる.