Barnesの第2補題

Barnesの第1補題 より,
$$\begin{array}{rl}\frac{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)\mathrm{\Gamma }(a+n)\mathrm{\Gamma }(b+n)}{\mathrm{\Gamma }(c+n)}& =\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(n-s)\mathrm{\Gamma }(c-a-b-s)ds\end{array}$$
である. 両辺に$\frac{(d{)}_n}{(e{)}_n}$を掛けて足し合わせると,
$$\begin{array}{rl}& \frac{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(c)}{}_3{F}_2[\begin{array}{c}a,b,d\\ c,e\end{array};1]\\ & =\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-a-b-s)\mathrm{\Gamma }(-s){}_2{F}_1[\begin{array}{c}d,-s\\ e\end{array};1]ds\\ & =\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-a-b-s)\mathrm{\Gamma }(-s)\frac{\mathrm{\Gamma }(e)\mathrm{\Gamma }(e-d+s)}{\mathrm{\Gamma }(e-d)\mathrm{\Gamma }(e+s)}ds\end{array}$$
より,
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(e-d+s)\mathrm{\Gamma }(c-a-b-s)\mathrm{\Gamma }(-s)}{\mathrm{\Gamma }(e+s)}ds\\ & =\frac{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(e-d)}{\mathrm{\Gamma }(c)\mathrm{\Gamma }(e)}{}_3{F}_2[\begin{array}{c}a,b,d\\ c,e\end{array};1]\end{array}$$
ここで, $d=c$とすると,
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(e-c+s)\mathrm{\Gamma }(c-a-b-s)\mathrm{\Gamma }(-s)}{\mathrm{\Gamma }(e+s)}ds\\ & =\frac{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(e-c)}{\mathrm{\Gamma }(c)\mathrm{\Gamma }(e)}{}_2{F}_1[\begin{array}{c}a,b\\ e\end{array};1]\\ & =\frac{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(e-c)\mathrm{\Gamma }(e-a-b)}{\mathrm{\Gamma }(c)\mathrm{\Gamma }(e-a)\mathrm{\Gamma }(e-b)}\end{array}$$
よって, $c\mapsto 1-d+a+b$として,
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(d+e-a-b-1+s)\mathrm{\Gamma }(1-d-s)\mathrm{\Gamma }(-s)}{\mathrm{\Gamma }(e+s)}ds\\ & =\frac{\mathrm{\Gamma }(1-d+a)\mathrm{\Gamma }(1-d+b)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(d+e-a-b-1)\mathrm{\Gamma }(e-a-b)}{\mathrm{\Gamma }(e-a)\mathrm{\Gamma }(e-b)\mathrm{\Gamma }(1-d+a+b)}\end{array}$$
改めて, $c=d+e-a-b-1$とすれば,
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c+s)\mathrm{\Gamma }(1-d-s)\mathrm{\Gamma }(-s)}{\mathrm{\Gamma }(e+s)}ds\\ & =\frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(c)\mathrm{\Gamma }(1-d+a)\mathrm{\Gamma }(1-d+b)\mathrm{\Gamma }(1-d+c)}{\mathrm{\Gamma }(e-a)\mathrm{\Gamma }(e-b)\mathrm{\Gamma }(e-c)}\end{array}$$
を得る.