# 一个简单的对数积分

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

$\Large\mathbf{Solution:}$

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

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

\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*}

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

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

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

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