44th Putnam 1983 Problem B5 Define ||x|| as the distance from x to the nearest integer. Find lim n→∞ 1/n ∫ 1 n ||n/x|| dx. You may assume that ∏ 1 ∞ 2n/(2n-1) 2n/(2n+1) = π/2. Solution
: ln(4/π). Note that on the interval (2n/2n, 2n/(2n-1) ) the integrand is n - n/x; on (2n/(2n-1), 2n/(2n-2) ) it is n/x - (n- 1); on (2n/(2n-2), 2n/(2n-3) ) it is (n-1) - n/x; ... ; on (2n/5, 2n/4) it is n/x - 2; on (2n/4, 2n/3) it is 2 - n/x; and on (2n/3, 2n/2) it is n/x - 1. Integrating the constant term over the interval (2n/(2r+1), 2n/2r) gives - r (2n/2r - 2n/(2r+1) ) = - n/(2r + 1). Integrating it over the adjacent interval (2n/(2r+2), 2n/(2r+1) ) gives + (r + 1) (2n/(2r+1) - 2n/(2r+2) ) = n/(2r + 1). So over the entire interval (1, n) the constant terms integrate to zero. Integrating the 1/x term over (2n/(2r+1), 2n/2r) gives n ln( (2r+1)/2r ), whilst over (2n/2r, 2n/(2r-1) ) gives - n ln( 2r/(2r-1) ). So over the combined interval (2n/(2r+1), 2n/(2r-1) ) we get - n ln ( 2r/(2r-1) 2r/(2r+1) ). Summing we get that integral over (2n/(2n-1), 2n/3) is - n ln P n , where P n = ∏ 2 n-1 2r/(2r-1) 2r/(2r+1). That leaves the intervals at each end which integrate to n ln 3/2 and - n ln(2n/(2n-1) ). So multiplying by 1/n and taking the limit we get - ln(3/4 π/2) + ln 3/2 = ln(4/π). 44th Putnam 1983 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001