Barnesの第1補題

留数定理により,
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-s)\mathrm{\Gamma }(d-s)ds\\ & =\sum _{0\le n}{\mathrm{Res}}_{s=c+n}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-s)\mathrm{\Gamma }(d-s)\\ & +\sum _{0\le n}{\mathrm{Res}}_{s=d+n}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-s)\mathrm{\Gamma }(d-s)\\ & =\sum _{0\le n}\frac{(-1{)}^n}{n!}\mathrm{\Gamma }(a+c+n)\mathrm{\Gamma }(b+c+n)\mathrm{\Gamma }(d-c-n)\\ & +\sum _{0\le n}\frac{(-1{)}^n}{n!}\mathrm{\Gamma }(a-d+n)\mathrm{\Gamma }(b-d+n)\mathrm{\Gamma }(c-d-n)\\ & =\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(d-c){}_2{F}_1[\begin{array}{c}a+c,b+c\\ 1+c-d\end{array};1]+\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+d)\mathrm{\Gamma }(c-d){}_2{F}_1[\begin{array}{c}a+d,b+d\\ 1+d-c\end{array};1]\\ & =\frac{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(d-c)\mathrm{\Gamma }(1+c-d)\mathrm{\Gamma }(1-a-b-c-d)}{\mathrm{\Gamma }(1-a-d)\mathrm{\Gamma }(1-b-d)}\\ & +\frac{\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+d)\mathrm{\Gamma }(c-d)\mathrm{\Gamma }(1+d-c)\mathrm{\Gamma }(1-a-b-c-d)}{\mathrm{\Gamma }(1-a-c)\mathrm{\Gamma }(1-b-c)}\\ & =\frac{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}{\mathrm{\Gamma }(a+b+c+d)}\frac{\sin \pi (a+d)\sin \pi (b+d)-\sin \pi (a+c)\sin \pi (b+c)}{\sin \pi (d-c)\sin \pi (a+b+c+d)}\\ & =\frac{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}{\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
となって示される.

Mellin逆変換公式より,
$$\begin{array}{rl}F(s)& ={\int }_0^{\mathrm{\infty }}{x}^{s-1}f(x)dx\\ G(s)& ={\int }_0^{\mathrm{\infty }}{x}^{s-1}g(x)dx\end{array}$$
とするとき,
$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}f(x)g(x)dx& =\frac{1}{2\pi i}{\int }_0^{\mathrm{\infty }}g(x){\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}F(t)x^{-t}dtdx\\ & =\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}F(t)G(1-t)dt\end{array}$$
が成り立つ. よって,
$$\begin{array}{rl}f(x)& =\frac{x^b}{(1+x{)}^a}\\ g(x)& =\frac{x^c}{(1+x{)}^d}\end{array}$$
とすると, ベータ積分より,
$$\begin{array}{r}F(s)=\frac{\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(a-b-s)}{\mathrm{\Gamma }(a)}\\ G(s)=\frac{\mathrm{\Gamma }(d+s)\mathrm{\Gamma }(c-d-s)}{\mathrm{\Gamma }(c)}\end{array}$$
であるから,
$$\begin{array}{rl}& \frac{\mathrm{\Gamma }(b+d+1)\mathrm{\Gamma }(a+c-b-d-1)}{\mathrm{\Gamma }(a+c)}\\ & ={\int }_0^{\mathrm{\infty }}\frac{x^{b+d}}{(1+x{)}^{a+c}}dx\\ & =\frac{1}{2\pi i\mathrm{\Gamma }(a)\mathrm{\Gamma }(c)}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(a-b-s)\mathrm{\Gamma }(1+d-s)\mathrm{\Gamma }(c-d-1+s)ds\end{array}$$
を得る. よって, 文字を適当に置き換えることによって,
$$\begin{array}{rl}\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-s)\mathrm{\Gamma }(d-s)ds& =\frac{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}{\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
が得られる.