一个简单的对数积分

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


\(\Large\mathbf{Solution:}\)
方法一:考虑含参积分

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

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

往下积分有理积分形式.


方法二:

\[\begin{align*} \int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x^{2}}\mathrm{d}x&=\sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n+1}}{n}\int_{0}^{1}\frac{x^{2n}}{1+x^{2}}\mathrm{d}x\\&=\frac{1}{4}\sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n+1}}{n}\left ( \psi _{0}\left ( \frac{n}{2}+\frac{3}{4} \right )-\psi _{0}\left ( \frac{n}{2}+\frac{1}{4} \right ) \right ) \end{align*}\]

接下来就涉及到了 Euler Sum的计算.上述两种方法都过于麻烦,现在我们看一种更为简单的方法.


方法三:做代换 \(x=\tan t\),我们有

\[\int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x^{2}}\mathrm{d}x=\int_{0}^{\frac{\pi }{4}}\ln\left ( 1+\tan^{2}t \right )\mathrm{d}t=-2\int_{0}^{\frac{\pi }{4}}\ln\cos t\mathrm{d}t \]

因为

\[\int_{0}^{\frac{\pi }{4}}\ln\cos t\mathrm{d}t=\frac{1}{2}\left ( \mathbf{G}-\frac{\pi }{2}\ln 2 \right ) \]

所以

\[\Large\boxed{\displaystyle \int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x^{2}}\mathrm{d}x=\color{blue}{\frac{\pi }{2}\ln 2-\mathbf{G}}} \]

posted @ 2016-05-11 19:52  Renascence_5  阅读(439)  评论(0编辑  收藏