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.

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998