<s>[INST] For what $(a,b) \in R^+$ does $\int^\infty_b (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}})dx$ converge? For what pairs $(a,b) \in R^+$ does this integral converge?
$$
\int\limits^{\infty}_{b} \left (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}} \right)dx
$$
 [/INST] The integral converges iff $a = b$.
Let's rewrite the expression using big-O notation. So
$$\sqrt{x+a}-\sqrt{x} = x^{1/2}(\sqrt{1+a/x}-1) = x^{1/2}(1+a/2x+O(x^{-2})),$$
Therefore
$$\sqrt{\sqrt{x+a}-\sqrt{x}} = x^{1/4}(a/4x+O(x^{-2}))$$
and similarly
$$\sqrt{\sqrt{x}-\sqrt{x-b}} = x^{1/4}(b/4x+O(x^{-2})).$$
Hence
$$
\int\limits^{\infty}_{b} \left (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}} \right) dx = \int\limits^{\infty}_{b} \left (x^{1/4}((a-b)/4x + O(x^{-2})\right)dx = \int\limits^{\infty}_{b} \left (x^{1/4}O(x^{-2})\right)dx + \frac{(a-b)}{4}\int\limits^{\infty}_{b} \left (x^{-3/4}\right)dx = \int\limits^{\infty}_{b} \left (O(x^{-7/4})\right)dx + \frac{(a-b)}{4}\int\limits^{\infty}_{b} \left (x^{-3/4}\right)dx.
$$
The first integral converges while the second integral doesn't unless $a=b$.
NOTE $O(x^n)$ means bounded by $ax^n$ ($a$ is a constant) and thus $(1+x)^{1/2}=1+x/2+O(x^2)$ for $|x| < 1$.
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