HMMT 二月 2011 · ALGCALC 赛 · 第 11 题

英文解析

11. Evaluate dx .
    1 xn − 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 xx =1 x =1
    x =1
    (ln x )n
    But ln(1) = 0, and lim = 0 for all n > 0. Sox →∞ 2010
    ∫ ∫x
    ∞ ∞
    n n − 1
    (ln x ) n (ln x )
    dx = dx
    2011 2011
    x 2010 xx =1 x =1
    It follows that
    ∫ ∫
    ∞ ∞
    (ln x ) n ! 1 n !ndx = dx =
    2011 n 2011 n +1
    x 2010 x 2010
    x =1 x =1
    2011!
    So the answer is .
    2012
    2010
    Algebra & Calculus Individual Test 2