HMMT 二月 2008 · CALC 赛 · 第 10 题
HMMT February 2008 — CALC Round — Problem 10
题目详情
- [ 8 ] Evaluate the integral ln x ln(1 − x ) dx . 0 1
解析
- [ 8 ] Evaluate the integral ln x ln(1 − x ) dx . 0 2 2 3 π x x Answer: 2 − We have the MacLaurin expansion ln(1 − x ) = − x − − − · · · . So 6 2 3 ∫ ∫ ∫ ∞ ∞ 1 1 1 n ∑ ∑ x 1 n ln x ln(1 − x ) dx = − ln x dx = − x ln x dx. n n 0 0 0 n =1 n =1 Using integration by parts, we get ∣ 1 ∫ ∫ 1 ∣ 1 n +1 n x ln x x 1 ∣ n x ln x dx = − dx = − . ∣ 2 n + 1 ∣ n + 1 ( n + 1) 0 0 0 n (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 is 2 n =0 n 6 2 π equal to 2 − . 6 3