<s>[INST] An improper integral involving nested square roots I have:
$$I_{a,b}= \int_b^{ + \infty } \left( \sqrt {\sqrt {x + a}  - \sqrt x }  - \sqrt {\sqrt x  - \sqrt {x - b} }  \right)dx$$
with $a>0$ and $b>0$.
I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.
 [/INST] We have:
$$\sqrt{x+a}-\sqrt{x}=\frac{a}{\sqrt{x}+\sqrt{x+a}},\qquad \sqrt{x}-\sqrt{x-b}=\frac{b}{\sqrt{x}+\sqrt{x-b}}$$
and:
$$f_{a,b}(x)=\sqrt{\sqrt{x+a}-\sqrt{x}}-\sqrt{\sqrt{x}-\sqrt{x-b}}=\frac{\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}}{\frac{\sqrt{a}}{\sqrt{\sqrt{x}+\sqrt{x+a}}}+\frac{\sqrt{b}}{\sqrt{\sqrt{x}+\sqrt{x-b}}}}$$
hence $f_{a,b}(x)$ behaves like
$$ x^{1/4}\left(\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}\right) $$
as $x\to +\infty$. If $a\neq b$, $I_{a,b}$ cannot converge, since $x^{-1/4}\not\in L^1((1,+\infty))$. 
If $a=b$ then $f_{a,b}(x)$ behaves like
$$ x^{1/4}\left(\frac{1}{\sqrt{x}+\sqrt{x+a}}-\frac{1}{\sqrt{x}+\sqrt{x-a}}\right)=x^{1/4}\left(\frac{\sqrt{x-a}-\sqrt{x+a}}{(\sqrt{x}+\sqrt{x+a})(\sqrt{x}+\sqrt{x-a})}\right)$$
that is $\Theta(x^{1/4-3/2})$ as $x\to +\infty$, hence $I_{a,b}$ converges.
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