41st Putnam 1980 Problem A3 Find ∫ 0 π/2 f(x) dx, where f(x) = 1/(1 + tan √2 x). Solution
: π/4. This involves a trick. f(π/2 - x) = 1/(1 + cot √2 x) = tan √2 x/(1 + tan √2 x) = 1 - f(x). Hence ∫ 0 π/2 f(x) dx = ∫ 0 π/4 f(x) dx + ∫ π/4 π/2 f(x) dx = ∫ 0 π/4 f(x) dx + ∫ 0 π/4 f(π/2 - x) dx = ∫ 0 π/4 f(x) dx + ∫ 0 π/4 (1 - f(x) ) dx = ∫ 0 π/4 dx = π/4. Note that the √2 is irrelevant - the argument works for any exponent. 41st Putnam 1980 © John Scholes jscholes@kalva.demon.co.uk 10 Dec 1999 Last corrected/updated 21 Nov 02