MHB What pairs of positive real numbers make the given integral converge?

[Ackbach]

Here is this week's POTW:

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

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[Ackbach]

Re: Problem Of The Week # 213 - April 26, 2016

This was Problem A-2 in the 1995 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

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
\begin{align*}
\sqrt{x+a}-\sqrt{x} &= x^{1/2}(\sqrt{1+a/x} - 1) \\
&= x^{1/2}(1 + a/2x + O(x^{-2})),
\end{align*}
hence
\[
\sqrt{\sqrt{x+a} - \sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))
\]
and similarly
\[
\sqrt{\sqrt{x} - \sqrt{x-b}} = x^{1/4} (b/4x + O(x^{-2})).
\]
Hence the integral we're looking at is
\[
\int_{b}^{\infty} x^{1/4} ((a-b)/4x + O(x^{-2}))\,dx.
\]
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).