43rd Putnam 1982 Problem A3 Evaluate ∫ 0 ∞ (tan -1 (πx) - tan -1 x) / x dx. Solution
: π/2 ln π. ∫ 0 ∞ (tan -1 (πx) - tan -1 x) / x dx = ln x (tan -1 πx - tan -1 x) | 0 ∞ - ∫ 0 ∞ ( π(1 + π 2 x 2 ) - 1/(1 + x 2 ) ) ln x dx. Putting y = πx, ∫ 0 ∞ ( π(1 + π 2 x 2 ) dx = ∫ 0 ∞ (ln y - ln π)/(1 + y 2 ) dy. So - ∫ 0 ∞ ( π(1 + π 2 x 2 ) - 1/(1 + x 2 ) ) ln x dx = ln π ∫ 0 ∞ dy/(1 + y 2 ) = ln π tan -1 y | 0 ∞ = π/2 ln π. It remains to show that the first term is zero. The Taylor series for tan -1 x = x + O(x 2 ), so the value at the lower limit is the limit as x → 0 of ( (π - 1) x + O(x 2 ) ) ln x. But x ln x → 0 as x → 0, so the lower limit gives zero. We also have the expansion tan -1 x = π/2 - 1/x + O(1/x 2 ). So the value at the upper limit is the limit of ( 1/x (1 - 1/π) + O(1/x 2 ) ) ln x as x → ∞, which is zero since (ln x)/x → 0. 43rd Putnam 1982 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001