6楼: Originally posted by alober at 2017-06-26 22:15:36
I(n) = \int_{0}^{+\infty}\frac{\ln^nx}{1+x^2}dx \mathop{=}^{x \to e^{-x}} \int_{-\infty}^{+\infty}\frac{(-x)^n}{1+e^{-2x}}e^{-x}dx = \begin{cases}0 & 2 \nmid n\\(-1)^n \cdot 2\int_{0}^{+\infty}\f ...