8楼: Originally posted by
Edstrayer at 2014-11-05 03:35:59
\int_1^{+\infty}\frac{(x-1)^n}{x^m}dx=\int_1^{+\infty}\frac{\sum\limits_{i=0}^n{n \choose i}(-1)^ix^{n-i}}{x^m}dx
\int_1^{+\infty}\frac{(x-1)^n}{x^m}dx=\sum\limits_{i=0}^n(-1)^i{n \choose i}\int_1^{ ...