计算下面两个积分的比值：
\[\Large\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+t^{4}}}\mathrm{d}t~,~\int_{0}^{1}\frac{1}{\sqrt{1-t^{4}}}\mathrm{d}t\]
\(\Large\mathbf{Solution:}\)
由
\[\begin{align*} \int_{0}^{1}\frac{1}{\sqrt{1+t^{4}}}\mathrm{d}t &= \int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left ( 1+t^{2} \right )^{2}-2t^{2}}}\\ &=\frac{1}{2}\int_{0}^{1}\frac{1}{\sqrt{1-\dfrac{1}{2}\left ( \dfrac{2t}{1+t^{2}} \right )^{2}}}\frac{2}{1+t^{2}}\mathrm{d}t\\ \left ( t=\tan\frac{\theta }{2} \right )\ \ &=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\mathrm{d}\theta }{\sqrt{1-\dfrac{1}{2}\sin^{2}\theta }}=\frac{1}{2}\mathbf{K}\left ( \frac{1}{\sqrt{2}} \right ) \end{align*}\]
和
\[\begin{align*} \int_{0}^{1}\frac{1}{\sqrt{1-t^{4}}}\mathrm{d}t&=\int_{0}^{\frac{\pi }{2}}\frac{\mathrm{d}\theta }{\sqrt{1+\cos^{2}\theta }} \ \ \left ( t=\cos \theta \right )\\ &=\int_{0}^{\frac{\pi }{2}}\frac{\mathrm{d}\theta }{\sqrt{2-\sin^{2}\theta }}\\ &=\frac{1}{\sqrt{2}}\mathbf{K}\left ( \frac{1}{\sqrt{2}} \right ) \end{align*}\]
我们可以得到
\[\Large\boxed{\displaystyle {\frac{\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1-x^{4}}}\mathrm{d}x}{\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x^{4}}}\mathrm{d}x}}=\frac{\dfrac{1}{\sqrt{2}}\mathbf{K}\left ( \dfrac{1}{\sqrt{2}} \right )}{\dfrac{1}{2}\mathbf{K}\left ( \dfrac{1}{\sqrt{2}} \right )}=\color{blue}{\sqrt {2}}}\]
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