双阶乘的估计
双阶乘除式的一个估计
\[\boxed{\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}} \sim \sqrt {n\pi } .}\]
证明：
\begin{align*}\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}} =\frac{{{{\left[ {\left( {2n} \right)!!} \right]}^2}}}{{\left( {2n} \right)!}} =\frac{{{2^{2n}}{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}} \sim \frac{{{2^{2n}}{{\left[ {{{\left( {\frac{n}{e}} \right)}^n}\sqrt {2\pi n} } \right]}^2}}}{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}\sqrt {4\pi n} }} = \frac{{{2^{2n}}{{\left[ {{{\left( {\frac{n}{e}} \right)}^n}\sqrt {2\pi n} } \right]}^2}}}{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}\sqrt {4\pi n} }} = \sqrt {n\pi }. \end{align*}
由此可对如下一个关于$\sin x$的整数次幂在$[-\frac{\pi}{2},\frac{\pi}{2}]$上的积分进行数值计算：
\[\int_0^{\frac{\pi }{2}} {{{\sin }^n}xdx} = \left\{ \begin{array}{l}\frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} \sim \frac{1}{2}\sqrt{\frac{\pi }{m}} ,n = 2m\\\frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \sim \frac{{\sqrt {m\pi } }}{{2m + 1}},n = 2m + 1\end{array} \right.,m \in {N_ + }.\]
特殊地，我们有：
\[0.250037 \approx I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\sin }^{100}}xdx} = 2\int_0^{\frac{\pi }{2}} {{{\sin }^{100}}xdx} = \frac{{99!!}}{{100!!}}\pi \sim \frac{1}{{\sqrt {50\pi } }}\pi = \frac{1}{5}\sqrt {\frac{\pi }{2}} = 0.25003696.\]
由上可知，此误差是很小的。
