用高斯算术几何平均数计算gamma(1/4)

永远

\(\displaystyle\begin{align}\frac{\pi }{{2M\left( {1,\sqrt 2 } \right)}} &= \int_0^{\frac{\pi }{2}} {\frac{{d\theta }}{{\sqrt {{{\cos }^2}\theta  + 2{{\sin }^2}\theta } }}} \\
&= \int_0^{\frac{\pi }{2}} {\frac{{d\theta }}{{\sqrt {1 + {{\sin }^2}\theta } }}} \\
&= \int_0^1 {\frac{{d(\arcsin t)}}{{\sqrt {1 + {t^2}} }}} \\
&= \int_0^1 {\frac{{dt}}{{\sqrt {1 - {t^2}} \sqrt {1 + {t^2}} }}} \\
&= \int_0^1 {\frac{{dt}}{{\sqrt {1 - {t^4}} }}} \\
&= \int_0^1 {\frac{{d(\sqrt[4]{x})}}{{\sqrt {1 - x} }}} \\
&= \int_0^1 {\frac{{\frac{1}{4}{x^{ - \frac{3}{4}}}dx}}{{\sqrt {1 - x} }}} \\
&= \frac{1}{4}\int_0^1 {{x^{ - \frac{3}{4}}}{{(1 - x)}^{ - \frac{1}{2}}}} dx\\
&= \frac{1}{4}B(\frac{1}{4},\frac{1}{2})\\
&= \frac{1}{4}\frac{{\Gamma (\frac{1}{4})\Gamma (\frac{1}{2})}}{{\Gamma (\frac{1}{4} + \frac{1}{2})}}\\
&= \frac{{\sqrt \pi  }}{4}\frac{{\Gamma (\frac{1}{4})}}{{\Gamma (\frac{3}{4})}}\\
&= \frac{{\sqrt \pi  }}{4}\frac{{{\Gamma ^2}(\frac{1}{4})}}{{\sqrt 2 \pi }}\\
&= \frac{{{\Gamma ^2}(\frac{1}{4})}}{{4\sqrt {2\pi } }}
\end{align}\)


\(\displaystyle\boxed{\frac{\pi }{{2M\left( {1,\sqrt 2 } \right)}} = \frac{{{\Gamma ^2}(\frac{1}{4})}}{{4\sqrt {2\pi } }} \Longrightarrow \Gamma (\frac{1}{4}) = \sqrt {\frac{{2\pi \sqrt {2\pi } }}{{M\left( {1,\sqrt 2 } \right)}}} } \)