HMMT 二月 2008 · CALC 赛 · 第 10 题

英文解析

10. [ 8 ] Evaluate the integral ln x ln(1 − x ) dx .
    2 2 30
    π x x
    Answer: 2 − We have the Mac laurin expansion ln(1 − x ) = − x − − − · · · . So
    6 2 3
    ∫ ∫ ∫
    ∞ ∞
    1 1 1
    ∑ ∑nx 1
    ln x ln(1 − x ) dx = − ln x dx = − x ln x dx.nn n
    0 0 0
    n =1 n =1
    Using integration by parts, we get
    ∣
    ∫ ∫1
    1 ∣ 1
    n +1 nx ln x x 1
    ∣
    x ln x dx = − dx = − .n
    ∣
    n + 1 ∣ n + 1 ( n + 1)2
    0 0
    n 0
    (We used the fact that lim x ln x = 0 for n > 0, which can be proven using l’Hˆ opital’s rule.)
    x → 0
    Therefore, the original integral equals to
    ( )
    ∞ ∞
    ∑ ∑
    1 1 1 1 = − − .
    2 2
    n ( n + 1) n n + 1 ( n + 1)
    n =1 n =1
    ∑
    ∞2
    1 π
    Telescoping the sum and using the well-known identity = , we see that the above sum isn =02
    n 6
    π2
    equal to 2 − .
    36