Convergence of $\int_b^\infty \left(\sqrt{\sqrt{x+a} - \sqrt{x}} - \sqrt{\sqrt{x} - \sqrt{x-b}}\right)dx$

Question

For what pairs $(a, b)$ of positive real numbers does the improper integral \begin{equation} \int_b^\infty \left(\sqrt{\sqrt{x+a} - \sqrt{x}} - \sqrt{\sqrt{x} - \sqrt{x-b}}\right)dx \end{equation}

converge.

I conjecture that $a = b$, but I'm not sure how to approach the proof. Some hints would be greatly appreciated!

Answer 1

1. If you are aware of the Big-O notation, then using the Taylor's Theorem, we get

$$\sqrt{1+\varepsilon} = 1 + \frac{1}{2}\varepsilon + \mathcal{O}(\varepsilon^2) \quad \text{as} \quad \varepsilon \to 0.$$

As such,

$$ \sqrt{x+a} = \sqrt{x} \sqrt{1 + \frac{a}{x}} = \sqrt{x} + \frac{a}{2\sqrt{x}} + \mathcal{O}\left(\frac{1}{x^{3/2}}\right) \quad \text{as} \quad x \to \infty, $$

and similarly

$$ \sqrt{x-b} = \sqrt{x} - \frac{b}{2\sqrt{x}} + \mathcal{O}\left(\frac{1}{x^{3/2}}\right) \quad \text{as} \quad x \to \infty. $$

Plugging this back and applying the same estimate again,

$$ \sqrt{\sqrt{x+a} - \sqrt{x}} = \sqrt{\frac{a}{2\sqrt{x}} + \mathcal{O}\left(\frac{1}{x^{3/2}}\right)} = \sqrt{\frac{a}{2}} \frac{1}{x^{1/4}} + \mathcal{O}\left(\frac{1}{x^{5/4}}\right) $$

and

$$ \sqrt{\sqrt{x} - \sqrt{x-b}} = \sqrt{\frac{b}{2}} \frac{1}{x^{1/4}} + \mathcal{O}\left(\frac{1}{x^{5/4}}\right). $$

Plugging this back, the integrand is now asymptotically

$$ \left( \sqrt{\frac{a}{2}} - \sqrt{\frac{b}{2}} \right) \frac{1}{x^{1/4}} + \mathcal{O}\left(\frac{1}{x^{5/4}}\right), $$

and for the integral of this over $[b, \infty)$ to converge, the first term must vanish. (The error term $\mathcal{O}(x^{-5/4})$ does not pose any issue to the integrability, because we know that $\int_{b}^{\infty} x^{-5/4} \, \mathrm{d}x$ is finite.) Therefore the integral converges if and only if $a = b$.

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2. If you are not used to analyzing the asymptotic behavior of a function, then so called "rationalization" technique will be helpful:

\begin{align*} \sqrt{x+a} - \sqrt{x} = (\sqrt{x+a} - \sqrt{x}) \cdot \frac{\sqrt{x+a} + \sqrt{x}}{\sqrt{x+a} + \sqrt{x}} = \frac{a}{\sqrt{x+a} + \sqrt{x}}, \end{align*}

and similarly we have

$$ \sqrt{x}-\sqrt{x-b} = \frac{b}{\sqrt{x} + \sqrt{x-b}}. $$

Plugging this back,

\begin{align*} &\sqrt{\sqrt{x+a} - \sqrt{x}} - \sqrt{\sqrt{x}-\sqrt{x-b}} \\ &= \frac{\sqrt{a}}{\sqrt{\sqrt{x+a} + \sqrt{x}}} - \frac{\sqrt{b}}{\sqrt{\sqrt{x} + \sqrt{x-b}}} \\ &= \frac{\sqrt{a \left(\sqrt{x} + \sqrt{x-b}\right)} - \sqrt{b\left(\sqrt{x+a} + \sqrt{x}\right)}}{\sqrt{\left(\sqrt{x+a} + \sqrt{x}\right)\left(\sqrt{x} + \sqrt{x-b}\right)}} \\ &= \frac{a\left(\sqrt{x} + \sqrt{x-b}\right) - b\left(\sqrt{x+a} + \sqrt{x}\right)}{\sqrt{\left(\sqrt{x+a} + \sqrt{x}\right)\left(\sqrt{x} + \sqrt{x-b}\right)}\left(\sqrt{a \left(\sqrt{x} + \sqrt{x-b}\right)} + \sqrt{b\left(\sqrt{x+a} + \sqrt{x}\right)}\right)}. \end{align*}

• If $a = b$, then the numerator becomes $$ a \left(\sqrt{x+a} - \sqrt{x-a}\right) = \frac{2a^2}{\sqrt{x+a} + \sqrt{x-a}} \sim \frac{a^2}{\sqrt{x}} \quad \text{as} \quad x \to \infty. $$

• If $a \neq b$, then the numerator is $\sim 2(a - b)\sqrt{x} $.

• In any cases, the denominator is $\sim 2\sqrt{2}\left(\sqrt{a} + \sqrt{b} \right) x^{3/4} $.

Combining these two observations, we know that the integrand is of order $x^{-1/4}$ if $ a\neq b$ and of order $x^{-5/4}$ if $a = b$. So the integral converges if and only if $ a = b$.