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