40th Putnam 1979

Given 0 < α < β, find lim_λ→0 ( ∫₀¹ (βx + α(1 - x) )^λ dx )^1/λ.

Solution

Answer: 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(x²) ), so the expression is (1 + 1/k (β ln β - α ln α)/(β - α) + O(1/k²) )^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