de Branges-Wilson積分

Dougallの${}_5{F}_4$の和公式の積分類似
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(1+\frac{a}{2}+s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c+s)\mathrm{\Gamma }(d+s)\mathrm{\Gamma }(b-a-s)\mathrm{\Gamma }(-s)}{\mathrm{\Gamma }(\frac{a}{2}+s)\mathrm{\Gamma }(1+a-c+s)\mathrm{\Gamma }(1+a-d+s)}ds\\ & =\frac{1}{2}\frac{\mathrm{\Gamma }(b)\mathrm{\Gamma }(c)\mathrm{\Gamma }(d)\mathrm{\Gamma }(b+c-a)\mathrm{\Gamma }(b+d-a)}{\mathrm{\Gamma }(1+a-c-d)\mathrm{\Gamma }(b+c+d-a)}\end{array}$$
において, $a\mapsto 2a,b\mapsto a+b,c\mapsto a+c,d\mapsto a+d,s\mapsto s-a$とすると,
$$\begin{array}{rl}\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(s+1)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c+s)\mathrm{\Gamma }(d+s)\mathrm{\Gamma }(a-s)\mathrm{\Gamma }(b-s)}{\mathrm{\Gamma }(s)\mathrm{\Gamma }(1-c+s)\mathrm{\Gamma }(1-d+s)}ds& =\frac{1}{2}\frac{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}{\mathrm{\Gamma }(1-c-d)\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
をガンマ関数の相反公式を用いて, これは
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a-s)\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b-s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-s)\mathrm{\Gamma }(c+s)\mathrm{\Gamma }(d-s)\mathrm{\Gamma }(d+s)}{\mathrm{\Gamma }(2s)\mathrm{\Gamma }(-2s)}ds(-\frac{\sin \pi (c-s)\sin \pi (d-s)}{\sin \pi s\cos \pi s\sin (c+d)})\\ & =2\frac{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)\mathrm{\Gamma }(c+d)}{\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
と書き換えられる. $s\mapsto -s$としたものを足し合わせて$\frac{1}{2}$倍すると,
$$\begin{array}{rl}\frac{1}{2}(\frac{\sin \pi (c+s)\sin \pi (d+s)}{\sin \pi s\cos \pi s\sin (c+d)}-\frac{\sin \pi (c-s)\sin \pi (d-s)}{\sin \pi s\cos \pi s\sin (c+d)})& =1\end{array}$$
であるから,
$$\begin{array}{rl}& \frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{\mathrm{\Gamma }(a-s)\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(b-s)\mathrm{\Gamma }(b+s)\mathrm{\Gamma }(c-s)\mathrm{\Gamma }(c+s)\mathrm{\Gamma }(d-s)\mathrm{\Gamma }(d+s)}{\mathrm{\Gamma }(2s)\mathrm{\Gamma }(-2s)}ds\\ & =2\frac{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)\mathrm{\Gamma }(c+d)}{\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
を得る. よって,
$$\begin{array}{rl}& \frac{1}{2\pi }{\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a-ix)\mathrm{\Gamma }(a+ix)\mathrm{\Gamma }(b-ix)\mathrm{\Gamma }(b+ix)\mathrm{\Gamma }(c-ix)\mathrm{\Gamma }(c+ix)\mathrm{\Gamma }(d-ix)\mathrm{\Gamma }(d+ix)}{\mathrm{\Gamma }(2ix)\mathrm{\Gamma }(-2ix)}dx\\ & =\frac{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)\mathrm{\Gamma }(c+d)}{\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
が成り立つ.

定理1より,
$$\begin{array}{rl}& \frac{1}{2\pi }{\int }_s^{\mathrm{\infty }}\mathrm{\Gamma }(a+it)\mathrm{\Gamma }(b+it)\mathrm{\Gamma }(c-it)\mathrm{\Gamma }(d-it)\frac{\mathrm{\Gamma }(a-2is-it)\mathrm{\Gamma }(b-2is-it)\mathrm{\Gamma }(c+2is+it)\mathrm{\Gamma }(d+2is+it)}{\mathrm{\Gamma }(-2is-2it)\mathrm{\Gamma }(2it+2is)\mathrm{\Gamma }(a+b-2is)\mathrm{\Gamma }(c+d+2is)}dt\\ & =\frac{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}{\mathrm{\Gamma }(a+b+c+d)}\end{array}$$
が得られるので, $s\mapsto -\mathrm{\infty }$とすると示される.