一个超几何函数类型的积分
\[\Large\displaystyle \int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x
\]
\(\Large\mathbf{Solution:}\)
易知
\[\int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x=\frac{1}{3}\int_{0}^{1}x^{-\frac{1}{3}}\left ( 1+x \right )^{\frac{1}{2}}\mathrm{d}x
\]
下面我们来看这个一般形式
\[\int_{0}^{u}y^{b -1}\left ( u-y \right )^{c-b-1}\left ( y+\frac{u}{x} \right )^{-a}\mathrm{d}t
\]
首先我们引入Beta函数
\[\mathrm{B}\left ( a,b \right )=\int_{0}^{1}t^{a-1}\left ( 1-t \right )^{b-1}\mathrm{d}t
\]
然后引入超几何函数 \(_{2}F_{1}\) 的定义
\[_{2}F_{1}\left ( a,b;c;x \right )=\frac{1}{\mathrm{B}\left ( b,c-b \right )}\int_{0}^{1}t^{b-1}\left ( 1-t \right )^{c-b-1}\left ( 1-tx \right )^{-a}\mathrm{d}t
\]
简单调整之后我们可以得到
\[\int_{0}^{1}t^{b}\left ( 1-t \right )^{c}\left ( 1-tx \right )^{a}\mathrm{d}t=\mathrm{B}\left ( b+1,c+1 \right )\, _{2}F_{1}\left (-a,b+1;b+c+2;x \right )
\]
做代换 \(y=tu~,~x\rightarrow -x\)后，我们有
\[\begin{align*}
&\int_{0}^{1}t^{b}\left ( 1-t \right )^{c}\left ( 1-tx \right )^{a}\mathrm{d}t=\int_{0}^{1}\left ( \frac{y}{u} \right )^{b}\left ( 1-\frac{y}{u} \right )^{c}\left ( 1+\frac{yx}{u} \right )^{a}\frac{1}{u}\mathrm{d}y\\
&=\left ( \frac{u}{x} \right )^{-a}u^{-b-c-1}\int_{0}^{u}y^{b}\left ( u-y \right )^{c}\left ( y+\frac{u}{x} \right )^{a}\mathrm{d}y
\end{align*}\]
然后做代换 \(b+1\rightarrow b~,~c+1\rightarrow c-b~,~a\rightarrow -a\) 我们有
\[\int_{0}^{u}y^{b -1}\left ( u-y \right )^{c-b-1}\left ( y+\frac{u}{x} \right )^{-a}\mathrm{d}t=\left ( \frac{u}{x} \right )^{a}u^{c-1}\mathrm{B}\left ( b,c-b \right )\, _{2}F_{1}\left ( a,b;c;-x \right )
\]
所以我们令 \(u=1~,~x=1~,~b=\dfrac{2}{3}~,~c=\dfrac{5}{3}~,~a=-\dfrac{1}{2}\)，可以得到
\[\int_{0}^{1}x^{-\frac{1}{3}}\left ( 1+x \right )^{\frac{1}{2}}\mathrm{d}x=\frac{3}{2}\, _{2}F_{1}\left ( -\frac{1}{2},\frac{2}{3};\frac{5}{3};-1 \right )
\]
所以
\[\Large\boxed{\displaystyle \int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x=\color{blue}{\frac{1}{2}\, _{2}F_{1}\left ( -\frac{1}{2},\frac{2}{3};\frac{5}{3};-1 \right )}}
\]
