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?
$$
\int\limits^{\infty}_{b} \left (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}} \right)dx
$$
 A: Let's deal with the first term in the difference. We use $(1+h)^{1/2}= 1 + h/2 +O(h^2)$ throughout.
$$(x+a)^{1/2}-x^{1/2} = x^{1/2}[(1+a/x)^{1/2} -1] = x^{1/2}[1+a/(2x) + O(1/x^2) -1] = x^{1/2}[a/(2x) + O(1/x^2)] = (a/2)(1/x^{1/2})[1 + O(1/x)].$$
Now take the square root of that to get
$$(a/2)^{1/2}(1/x^{1/4})[1 + O(1/x)].$$
Similarly, the second term in the difference is
$$(b/2)^{1/2}(1/x^{1/4})[1 + O(1/x)].$$
So we subtract to get
$$[(a/2)^{1/2}-(b/2)^{1/2}](1/x^{1/4})+ O(1/x^{5/4}).$$
In absolute value this will be on the order of $1/x^{1/4}$ if $a\ne b.$ Since $1/x^{1/4}$ is nonintegrable at $\infty,$ for the convergence we must have $a=b.$ And if we do have $a=b,$ the integrand is on the order of $O(1/x^{5/4}),$ which gives a convergent integral.
Answer: The integral converges iff $a=b.$
A: 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 $|x| < 1$.
