HMMT 二月 2011 · CALCCOMB 赛 · 第 10 题

HMMT February 2011 — CALCCOMB Round — Problem 10

解析

Evaluate dx . x 1 n − 1 d n ln x 2011! n Answer: By the chain rule, (ln x ) = . 2012 2010 dx x We calculate the definite integral using integration by parts: [ ] ∫ ∫ x = ∞ ∞ ∞ n n n − 1 (ln x ) (ln x ) n (ln x ) dx = − dx 2011 2010 2011 x − 2010 x − 2010 x x =1 x =1 x =1 n (ln x ) But ln(1) = 0, and lim = 0 for all n > 0. So x →∞ 2010 x ∫ ∫ ∞ ∞ n n − 1 (ln x ) n (ln x ) dx = dx 2011 2011 x 2010 x x =1 x =1 It follows that ∫ ∫ ∞ ∞ n (ln x ) n ! 1 n ! dx = dx = 2011 n 2011 n +1 x 2010 x 2010 x =1 x =1 2011! So the answer is . 2012 2010