[Largedisplaystyle int_{0}^{1}frac{lnleft ( 1+x^{2}
ight )}{1+x^{2}}mathrm{d}x
]
(Largemathbf{Solution:})
方法一：考虑含参积分
[mathcal{I}left ( alpha
ight )=int_{0}^{1}frac{lnleft ( 1+alpha x^{2}
ight )}{1+x^{2}}mathrm{d}x
]
则
[mathcal{I}'left ( alpha
ight )=int_{0}^{1}frac{x^{2}}{left (1+x^{2}
ight )left ( 1+alpha x^{2}
ight )}mathrm{d}x
]
往下积分有理积分形式.
方法二：
[egin{align*}
int_{0}^{1}frac{lnleft ( 1+x^{2}
ight )}{1+x^{2}}mathrm{d}x&=sum_{n=1}^{infty }frac{left ( -1
ight )^{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
ight )^{n+1}}{n}left ( psi _{0}left ( frac{n}{2}+frac{3}{4}
ight )-psi _{0}left ( frac{n}{2}+frac{1}{4}
ight )
ight )
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= an t)，我们有
[int_{0}^{1}frac{lnleft ( 1+x^{2}
ight )}{1+x^{2}}mathrm{d}x=int_{0}^{frac{pi }{4}}lnleft ( 1+ an^{2}t
ight )mathrm{d}t=-2int_{0}^{frac{pi }{4}}lncos tmathrm{d}t
]
因为
[int_{0}^{frac{pi }{4}}lncos tmathrm{d}t=frac{1}{2}left ( mathbf{G}-frac{pi }{2}ln 2
ight )
]
所以
[Largeoxed{displaystyle int_{0}^{1}frac{lnleft ( 1+x^{2}
ight )}{1+x^{2}}mathrm{d}x=color{blue}{frac{pi }{2}ln 2-mathbf{G}}}
]