HMMT 二月 2009 · CALC 赛 · 第 10 题

HMMT February 2009 — CALC Round — Problem 10

[ 8 ] Let a and b be real numbers satisfying a > b > 0. Evaluate ∫ 2 π 1 dθ. a + b cos( θ ) 0 Express your answer in terms of a and b .

解析

[ 8 ] Let a and b be real numbers satisfying a > b > 0. Evaluate ∫ 2 π 1 dθ. a + b cos( θ ) 0 Express your answer in terms of a and b . 2 π √ Answer: 2 2 a − b 3 Solution: Using the geometric series formula, we can expand the integral as follows: ( ) ∫ ∫ 2 2 π 2 π 1 1 b b 2 dθ = 1 + cos( θ ) + cos ( θ ) dθ a + b cos( θ ) a a a 0 0 ( ) ( ) ∫ n ∞ n 2 π iθ − iθ ∑ 1 b e + e = dθ a a 2 0 n =0 ( ) ( ) ∞ n 2 n 2 ∑ 2 π b n = dθ 2 2 n a a 2 n =0 ( ) 2 n 1 To evaluate this sum, recall that C = is the n th Catalan number. The generating function n n +1 n for the Catalan numbers is √ ∞ ∑ 1 − 1 − 4 x n C x = , n 2 x n =0 ) ∑ ( 2 n n 1 √ and taking the derivative of x times this generating function yields x = . Thus the n 1 − 4 x 2 π √ integral evaluates to , as desired. 2 2 a − b 4