两个Beta函数类型的积分及其一般形式

\[\Large\displaystyle \int_{0}^{1}\frac{\sqrt[4]{x\left ( 1-x \right )^{3}}}{\left ( 1+x \right )^{3}}\mathrm{d}x~~,~~\int_{0}^{1}\frac{\sqrt[3]{x\left ( 1-x \right )^{2}}}{\left ( 1+x \right )^{3}}\mathrm{d}x\]

\(\Large\mathbf{Solution:}\) 我们来计算更一般的情况：\[I\left ( m,n \right )=\int_{0}^{1}\frac{\sqrt[n]{x^{m}\left ( 1-x \right )^{n-m}}}{\left ( 1+x \right )^{3}}\mathrm{d}x\] 为了化为Beta函数，我们作如下变换\[t=\frac{2x}{1+x},~ ~ ~ 1-t=\frac{1-x}{1+x},~ ~ ~ \mathrm{d}t=\frac{2\mathrm{d}x}{\left ( 1+x \right )^{2}}\] 于是直接计算得：\[\begin{align*} \int_{0}^{1}\frac{\sqrt[n]{x^{m}\left ( 1-x \right )^{n-m}}}{\left ( 1+x \right )^{3}}\mathrm{d}x &=\int_{0}^{1}\left ( \frac{x}{1+x} \right )^{\frac{m}{n}}\left ( \frac{1-x}{1+x} \right )^{\frac{n-m}{n}}\frac{\mathrm{d}x}{\left ( 1+x \right )^{2}} \\ &=2^{-\frac{n+m}{n}}\int_{0}^{1}t^{\frac{m}{n}}\left ( 1-t \right )^{\frac{n-m}{n}}\mathrm{d}t \\ &=\frac{2^{-\frac{n+m}{n}}}{\Gamma \left ( 3 \right )}\Gamma \left ( \frac{m+n}{n} \right )\Gamma \left ( \frac{2n-m}{n} \right ) \\ &=2^{-\frac{2n+m}{n}}\cdot \frac{m}{n}\cdot \frac{n-m}{n}\cdot \Gamma \left ( \frac{m}{n} \right )\cdot \Gamma \left ( 1-\frac{m}{n}\right )\\ &=2^{-\frac{2n+m}{n}}\cdot \frac{m\left ( n-m \right )}{n^{2}}\cdot \frac{\pi }{\sin\left ( \dfrac{m\pi }{n} \right )} \end{align*}\] 所以原题即为：\[\Large\boxed{\begin{align*} I\left ( 1,4 \right )&=\color{Blue} {\frac{3\sqrt[4]{2}}{64}\pi}\\[15pt] I\left ( 1,3\right )&=\color{Blue} {\frac{\pi }{18}\frac{\sqrt[3]{4}}{\sqrt{3}}} \end{align*}}\]

转载于:https://www.cnblogs.com/Renascence-5/p/5483303.html