Study the series of $\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$ with respect to $a>0$ I have to study the series $\sum a_n$ with $a_n=\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$, with respect to $a>0$.
I have thought to use the asymptotic criterion for series.
In particular I can observe that:
$\sin{x}\sim x-\frac{x^3}{3!}$ and so: $\sin{(\sqrt[3]{x})}\sim \sqrt[3]{x}- \frac{x}{3!}$ for $x\to 0$.
Now this means that: $$\int_0^{\frac{1}{n^a}}\sqrt[3]{x}- \frac{x}{3!}\,dx=[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}]\rvert_0^{\frac{1}{n^a}}-[\frac{x^{2}}{2\cdot 3!}]\rvert_0^{\frac{1}{n^a}}=\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}$$ Now since $\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}$ it self is asymptotic to $\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$ then I can say:
$$a_n\sim \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}\sim \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$$
Now $\sum_{n=1}^{\infty} \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$ is convergent for $\frac{4}{3}a>1\iff a>\frac{3}{4}$ and so from asymptote criterion also the original series converges for this value of $a$.
I need a check of my attempt and in case there is something wrong can you tell me where and how can I correct this?
(I can't use the Lebesgue theore since it is not something that I know)

Here is a simpler proof of the equivalent: let $f(t) = \frac{\sin(t)}{t}$. It is well known that $|f(t)|\le 1$ for all $t\in{\mathbb R}$ and that $f(0)=1$. The substitution $x = \frac{y}{n^a}$ in the integral gives
\begin{equation}
a_n = \left(\frac{1}{n^a}\right)^{4/3}\int_0^1 f\left(\left(\frac{y}{n^a}\right)^{1/3}\right) y^{1/3} d y
\end{equation}
By Lebesgue's dominated convergence theorem, the above integral converges to
$\int_0^1 y^{1/3} d y = \frac{3}{4}$ when $n$ tends to $\infty$, hence
\begin{equation}
a_n \sim \frac{3}{4}\left(\frac{1}{n^a}\right)^{4/3}
\end{equation}
