40th Putnam 1979 Problem B2 Given 0 < α < β, find lim λ→0 ( ∫ 0 1 (βx + α(1 - x) ) λ dx ) 1/λ . Solution
: β β/(β - α) /( e α α/(β - α) ) Let t = βx + α(1 - x). Then the integral becomes ∫ α β t λ dt/(β - α) = 1/(1 + λ) (β λ+1 - a λ+1 )/(β - α). We evaluate the limits of 1/(1 + λ) λ and (β λ+1 - a λ+1 ) λ /(β - α) λ separately. Put k = 1/λ. The first limit is lim 1/(1 + 1/k) k = 1/e. To evaluate the second, note that β x = e x ln β = (1 + x ln β + O(x 2 ) ), so the expression is (1 + 1/k (β ln β - α ln α)/(β - α) + O(1/k 2 ) ) k , which tends to the limit exp( (β ln β - α ln α)/(β - α) ) as k tends to infinity, in other words, β β/(β - α) / α α/(β - α) . 40th Putnam 1979 © John Scholes jscholes@kalva.demon.co.uk 4 Dec 1999