13th Putnam 1953

Using sin x = 2 sin x/2 cos x/2 or otherwise, find ∫₀^π/2 ln sin x dx.

Solution

Let I = ∫₀^π/2 ln sin x dx. Making the suggested substitution, we get I = π/2 ln 2 + ∫₀^π/2 ln sin x/2 cos x/2 dx. Putting y = x/2 we get ∫₀^π/2 ln sin x/2 cos x/2 dx = 2 ∫₀^π/4 ln sin y + ln cos y dy. But cos y = sin(π/2 - y), so ∫₀^π/4 ln cos y dy = ∫_π/4^π/2 ln sin y dy. Hence I = π/2 ln 2 + 2I, so I = - π/2 ln 2.

13th Putnam 1953

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002