Putnam 1995 Problem A2 For what positive reals α, β does ∫ β ∞ √(√(x + α) - √x) - √(√x - √(x - β) ) dx converge? Solution
: it converges for α = β. Straightforward. Multiplying up shows that given positive k  β, take k such that 0  β. Then the integrand is greater than (√(αk) - √β)/(x 1/4 √2). But the integral of x -1/4 diverges (because x 3/4 tends to infinity as x tends to infinity), do the original integral does also. Similarly, if α  α. Then the integrand is less than (√α - √(βk)/(x 1/4 √2) and so the integral again diverges. If α = β, then (√(x + α) - √x) takes the same values on the interval [nα, nα + α] as (√x - √(x - β)) does on the interval [nα+α, nα+2α]. Thus the integral from α to nα equals the K + L where - K is the integral from 0 to α of √(√(x + α) - √x) and L is the integral from nα-α to nα of √(√(x + α) - √x). But the integrand √(√(x + α) - √x) tends to zero as x tends to infinity, so L tends to zero as n tends to infinity and the integral converges to K. Putnam 1995 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998