積分解説8

$$\begin{gathered} \begin{array}{rcl}{\int }_0^1{\int }_0^1\frac{1}{\sqrt{1-x^2}}\frac{1}{\sqrt{1-y^2}}\frac{dxdy}{\sqrt{1-x^2{y}^2}}& =& {\int }_{0\le u,0\le v, \\ u+v\le \frac{\pi }{2}}\sqrt{\frac{{\cos }^{2}u{\cos }^{2}v-{\sin }^{2}u{\sin }^{2}v}{({\cos }^{2}u-{\sin }^{2}v)({\cos }^{2}v-{\sin }^{2}u)}}dudv(x=\frac{\sin v}{\cos u},y=\frac{\sin u}{\cos v})\\ & =& {\int }_{0\le u,0\le v, \\ u+v\le \frac{\pi }{2}}\sqrt{\frac{(\cos u\cos v-\sin u\sin v)(\cos u\cos v+\sin u\sin v)}{((\cos u-\sin v)(\cos v+\sin u))((\cos u+\sin v)(\cos v-\sin u))}}dudv\\ & =& {\int }_{0\le |t|\le s\le \frac{\pi }{2}}\frac{1}{2}\frac{dsdt}{\sqrt{\cos s}\sqrt{\cos t}}(\text{加法定理},\text{和積公式},u+v=s,u-v=t)\\ & =& \frac{1}{2}{\left(\frac{1}{2}B(\frac{1}{2},\frac{1}{4})\right)}^2\\ & =& \frac{1}{2}{\left(\frac{1}{2}{2}^{\frac{1}{2}-1}B(\frac{1}{4},\frac{1}{4})\right)}^2(B(\frac{1}{2},z)=2^{2z-1}B(z,z))\\ & =& \frac{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}{16\pi }\end{array} \end{gathered}$$