求此函數積分？

01-05

非初等函數，有什麼非常規方法求呢

$int_{0}^{1}frac{ln x}{1+x^2}dx =int_{0}^{1}ln xsum_{n=0}^{infty}x^{2n}dx =sum_{n=0}^{infty}int_{0}^{1}x^{2n}ln xdx$

對於右邊的積分，用分部積分法可得

$int x^{2n}ln xdx =frac{1}{2n+1}int ln xdx^{2n+1} =frac{1}{2n+1}left( x^{2n+1}ln x-int x^{2n}dx ight) =frac{1}{2n+1}x^{2n+1}ln x-frac{1}{left(2n+1 ight)^2}x^{2n+1}$

再代入上下限，可得 $int_{0}^{1}x^{2n}ln xdx=frac{1}{left(2n+1 ight)^2}$

於是所求積分即為

$int_{0}^{1}frac{ln x}{1+x^2}dx =sum_{n=0}^{infty}int_{0}^{1}x^{2n}ln xdx =sum_{n=0}^{infty}frac{1}{left(2n+1 ight)^2} =frac{pi^2}{8}$

只要知道

$int_0^{1} x^n ln^m xmathrm{d}x=left(-1 ight)^mfrac{m!}{left(n+1 ight)^{m+1}}$

就知道如何往下做了吧 ~