来自贴吧的一个题目

\[\Large\displaystyle \int_{0}^{1}\frac{\ln\left ( x+\sqrt{x^{2}+1} \right )}{x}\mathrm{d}x \]


\(\Large\mathbf{Solution:}\)
注意到:

\[\int_{0}^{1}\frac{\ln\left ( x+\sqrt{x^{2}+1} \right )}{x}\mathrm{d}x=\int_{0}^{1}\frac{\mathrm{arcsinh} x}{x}\mathrm{d}x \]

分部即得:

\[\int_{0}^{1}\frac{\mathrm{arcsinh} x}{x}\mathrm{d}x=\mathrm{arcsinh} x\ln x\Bigg|_{0}^{1}-\int_{0}^{1}\frac{\ln x}{\sqrt{1+x^{2}}}\mathrm{d}x \]

\[\mathcal{I}\left ( \alpha \right )=\int_{0}^{1}\frac{x^{\alpha }}{\sqrt{1+x^{2}}}\mathrm{d}x \]

我们有

\[\begin{align*} \mathcal{I}\left ( \alpha \right )&=\int_{0}^{1}\frac{x^{\alpha }}{\sqrt{1+x^{2}}}\mathrm{d}x=\int_{0}^{1}x^{\alpha } \left ( 1+x^{2} \right )^{-\frac{1}{2}}\mathrm{d}x\\ &=\frac{1}{2}\int_{0}^{1}x^{\frac{\alpha }{2}-\frac{1}{2}}\left ( 1+x \right )^{-\frac{1}{2}}\mathrm{d}x\\ &=\frac{1}{1+\alpha }\, _{2}F_{1}\left ( \frac{1}{2},\frac{1+\alpha }{2};\frac{3+\alpha }{2};-1 \right ) \end{align*}\]

所以

\[\large\boxed{\displaystyle \begin{align*} \int_{0}^{1}\frac{\ln\left ( x+\sqrt{x^{2}+1} \right )}{x}\mathrm{d}x&=-\int_{0}^{1}\frac{\ln x}{\sqrt{1+x^{2}}}\mathrm{d}x=-\mathcal{I}'\left ( 0 \right )\\ &=\color{blue}{\mathrm{arcsinh}\left ( 1 \right )-\frac{1}{2}\, _{2}F_{1}^{\left ( 0,1,0,0 \right )}\left ( \frac{1}{2},\frac{1}{2};\frac{3}{2};-1 \right )}\\[8pt] &~~~\color{blue}{-\frac{1}{2}\, _{2}F_{1}^{\left ( 0,0,1,0 \right )}\left ( \frac{1}{2},\frac{1}{2};\frac{3}{2};-1 \right )}\\[8pt] &\approx 0.95520180648118 \end{align*}}\]

posted @ 2016-06-06 14:16  Renascence_5  阅读(436)  评论(0编辑  收藏