Putnam 1995/A2 solution

Putnam 1995

Problem A2

For what positive reals α, β does ∫_β^∞ √(√(x + α) - √x) - √(√x - √(x - β) ) dx converge?

Solution

Answer: it converges for α = β.

Straightforward.

Multiplying up shows that given positive k < 1, αk/(2√x) < (√(x + α) - √x) < α/(2√x) for all sufficiently large x.

So if α > β, take k such that 0 < k < 1 and αk > 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 < β, take k such that 0 < k < 1 and βk > 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