Putnam 1995, Problem A-2

For what pairs \( (a,b) \) of positive real numbers does the improper integral
\[
\int_b^\infty \left( \sqrt{x+a} - \sqrt{x} - \sqrt{x - \sqrt{x - b}} \right) dx
\]
converge?

Solution:
The integral converges iff \( a = b \). The easiest proof uses "big-O" notation and the fact that \( (1 + x)^{1/2} = 1 + x/2 + O(x^2) \) for \( |x| < 1 \). (Here \( O(x^2) \) means bounded by a constant times \( x^2 \).)  
So  
\[
\sqrt{x + a} - \sqrt{x} = x^{1/2} \left( \sqrt{1 + a/x} - 1 \right) = x^{1/2} \left( 1 + \frac{a}{2x} + O(x^{-2}) \right),
\]  
hence  
\[
\sqrt{\sqrt{x + a} - \sqrt{x}} = x^{1/4} \left( \frac{a}{4x} + O(x^{-2}) \right)
\]  
and similarly  
\[
\sqrt{\sqrt{x} - \sqrt{x - b}} = x^{1/4} \left( \frac{b}{4x} + O(x^{-2}) \right).
\]  
Hence the integral we're looking at is  
\[
\int_b^\infty x^{1/4} \left( \frac{a - b}{4x} + O(x^{-2}) \right) dx.
\]  
The term \( x^{1/4} O(x^{-2}) \) is bounded by a constant times \( x^{-7/4} \), whose integral converges. Thus we only have to decide whether \( x^{-3/4} (a - b)/4 \) converges. But \( x^{-3/4} \) has divergent integral, so we get convergence if and only if \( a = b \) (in which case the integral telescopes anyway).