I have to study the convergence of this integral:
$\int_{0}^{1} \frac{1}{\sqrt x - \sqrt[5]{x}} dx$
I rewrite it as:
$\int_{0}^{a} f (x) dx + \int_{a}^{1} f (x) dx$ (with $0<a<1$)
in order to deal with the problems in 0 and 1 separately.
For the first one, since $ \sqrt x = o(\sqrt [5] x )$ , for $x \rightarrow 0$, it should be right to say that $f(x) \sim \frac{1}{- \sqrt[5]{x}}$, so it converges
As for the other (here's my doubt), I apply a change of variable (t=1-x) in order to obtain:
$\int_{a}^{1} f (x) dx = \int_{0}^{1-a} \frac{1}{\sqrt{1-t} - \sqrt[5]{1-t}} dt$
Then, using the relation $(1+x)^{b} \sim 1 + b x $ (for $ x \rightarrow 0$), I rewrite f (t) as:
$\frac{1}{(1-\frac {1}{2}t) - (1--\frac {1}{5}t)} $
so that
$ \int_{0}^{1-a} \frac{1}{\sqrt{1-t} - \sqrt[5]{1-t}} dt \sim \int_{0}^{1-a} -\frac{10}{3t} dt$
which diverges (negatively)
My question: Am I overcomplicating the whole thing, especially the second integral? I feel there should be a faster way for it but at the moment I'm not able to find it.
We are now examining the following integral: $$ \int_{0}^{1} \frac {1}{{x}^{1 / 5} - {x}^{1 / 2}} \text {d} x. $$ To begin with, we let $x \mapsto {x}^{10}$. So ${x}^{1 / 5} \mapsto {x}^{2}$, ${x}^{1 / 2} \mapsto {x}^{5}$, and $\text {d} x \mapsto 10 {x}^{9} \text {d} x$; accordingly, $$ \int_{0}^{1} \frac {1}{{x}^{1 / 5} - {x}^{1 / 2}} \text {d} x = 10 \int_{0}^{1} \frac {{x}^{7}}{1 - {x}^{3}} \text {d} x. $$ Let $x = 1 - h$. As $x \to {1}_{-}$, $h \to {0}_{+}$, and $$ \begin{align} \frac {{x}^{7}}{1 - {x}^{3}} \sim \frac {1}{1 - {x}^{3}} & = \frac {1}{1 - {\left( 1 - h \right)}^{3}} \\ & = \frac {1}{1 - \left( 1 - 3 h + \mathcal {O} \left( {h}^{2} \right) \right)} = \frac {1}{3 h} \cdot \frac {1}{1 + \mathcal {O} \left( h \right)} \sim \frac {1}{3 h}. \end{align} $$ Since we know that $$ \int_{0}^{a} \frac {1}{x} \text {d} x $$ diverges for all $a > 0$, the given integral must diverge.