積分解説8

$$ \begin{eqnarray} \int_{0}^{1}\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \frac{1}{ \sqrt{1-y^2}} \frac{dxdy}{ \sqrt{1-x^2y^2}}&=&\int_{0 \leq u,0 \leq v,\\u+v \leq \frac{\pi}{2}} \sqrt{ \frac{\cos^2{u}\cos^2{v}-\sin^2{u}\sin^2{v}}{(\cos^2{u}-\sin^2{v})(\cos^2{v}-\sin^2{u})}}dudv \quad \left( x=\frac{\sin{v}}{\cos{u}},y=\frac{\sin{u}}{\cos{v}} \right) \\ &=& \int_{0 \leq u,0 \leq v,\\u+v \leq \frac{\pi}{2}} \sqrt{ \frac{(\cos{u}\cos{v}-\sin{u}\sin{v})(\cos{u}\cos{v}+\sin{u}\sin{v})}{\big((\cos{u}-\sin{v})(\cos{v}+\sin{u})\big)\big((\cos{u}+\sin{v})(\cos{v}-\sin{u})\big)}}dudv \\ &=& \int_{0 \leq|t| \leq s\leq\frac{\pi}{2}} \frac{1}{2}\frac{dsdt}{\sqrt{\cos{s}}\sqrt{\cos{t}}} \quad \left( \text{加法定理},\text{和積公式}, u+v=s,u-v=t \right) \\ &=& \frac{1}{2}\left(\frac{1}{2}B\left(\frac{1}{2},\frac{1}{4}\right)\right)^2 \\ &=& \frac{1}{2}\left(\frac{1}{2}2^{\frac{1}{2}-1}B\left(\frac{1}{4},\frac{1}{4}\right)\right)^2 \quad \left( B\left(\frac{1}{2},z\right)=2^{2z-1}B\left(z,z\right) \right) \\ &=& \frac{\Gamma^4\left(\frac{1}{4}\right)}{16\pi} \end{eqnarray} $$