Convergence of real integral I'm trying to analyze the convergence of following integral:
$$
\int_{0}^{1}\frac{dx}{\sqrt[3]{x(e^{x}-e^{-x})}}
$$
Currently not being able to get a hint on how to proceed to do it, any help is really appreciated.
So far I've tried:
1.
$$
\frac{dx}{\sqrt[3]{x(e^{x}-e^{-x})}} >= \frac{dx}{\sqrt[3]{x(e-e^{-1})}}
$$
But right expression converges, so it's not helping.
2.
$$
\frac{dx}{\sqrt[3]{x(e^{x}-e^{-x})}} = \frac{dx}{\sqrt[3]{2xsh(x))}}
$$
But I couldn't get anything useful.
3.
I would expect this behaves as other easiest function which I could test that converges/diverges easily but I'm a bit lost on the transformation needed.
Thanks!

EDIT: Solution


*Proposed:
$$
\sim \frac{1}{x^{\frac{2}{3}}}
$$
Then comparisson by limits:
$$
\lim_{x \to 0} \frac{\frac{1}{\sqrt[3]{x(e^{x}-e^{-x})}}}{\frac{1}{x^{\frac{2}{3}}}} = \sqrt[3]{\lim_{x \to 0}\frac{x}{(e^{x}-e^{-x})}} =  \frac{1}{\sqrt[3]{2}}
$$
Finally as the limit is $\neq 0, \neq \infty$, then it behaves as the proposed function which converges.
Thank you!
 A: $e^x - e^{-x} > 2x > x$ for $x> 0$ so we have the inequality
$$\frac{1}{\sqrt[3]{x(e^x-e^{-x})}} \leq \frac{1}{x^{\frac{2}{3}}}$$
Thus the integral
$$0 < \int_0^1 \frac{dx}{\sqrt[3]{x(e^x-e^{-x})}}  \leq \int_0^1 x^{-\frac{2}{3}}\:dx = 3$$
converges.
A: The problem being around $x=0$, compose Taylor series
$$e^x-e^{-x}=2 x+\frac{x^3}{3}+\frac{x^5}{60}+O\left(x^7\right)$$
$$x(e^x-e^{-x})=2 x^2+\frac{x^4}{3}+\frac{x^6}{60}+O\left(x^8\right)$$ Now, binomial expansion
$$\sqrt[3]{x(e^x-e^{-x})}=\sqrt[3]{2} x^{2/3}+\frac{x^{8/3}}{9\ 2^{2/3}}-\frac{x^{14/3}}{1620\
   2^{2/3}}+O\left(x^{20/3}\right)$$ Long division
$$\frac 1{\sqrt[3]{x(e^x-e^{-x})}}=\frac{1}{\sqrt[3]{2} x^{2/3}}-\frac{x^{4/3}}{18 \sqrt[3]{2}}+\frac{11
   x^{10/3}}{3240 \sqrt[3]{2}}+O\left(x^{16/3}\right)$$ Integration
$$\int\frac 1{\sqrt[3]{x(e^x-e^{-x})}}=\frac{3 x^{1/3}}{\sqrt[3]{2}}-\frac{x^{7/3}}{42 \sqrt[3]{2}}+\frac{11
   x^{13/3}}{14040 \sqrt[3]{2}}+O\left(x^{19/3}\right)$$ and this is alternating. So
$$\frac{125}{42 \sqrt[3]{2}}\approx 2.36220<\int_0^1\frac {dx}{\sqrt[3]{x(e^x-e^{-x})}}<\frac{292577}{98280 \sqrt[3]{2}}\approx 2.36283$$ while the exact value is $\approx 2.36279$.
