Convergence of improper integral $\int_{0}^{\infty} \frac{\sin(x)}{x^{1/3}} dx$

Question

I first tried to separate the integral from 0 to 1 and 1 to $\infty$ but it didn't work because the upper bound is divergent. Any ideas? Thanks in advance.

Answer 1

\begin{align*} \int_{0}^{\infty}\dfrac{\sin x}{x^{1/3}}dx&=\int_{0}^{1}\dfrac{\sin x}{x^{1/3}}dx+\int_{1}^{\infty}\dfrac{\sin x}{x^{1/3}}dx, \end{align*} where $\left|\dfrac{\sin x}{x^{1/3}}\right|\leq\dfrac{1}{x^{1/3}}$ on $[0,1]$ and hence the first improper integral converges absolutely.

For the second improper integral, note that \begin{align*} \int_{1}^{\infty}\dfrac{\sin x}{x^{1/3}}dx&=-\dfrac{\cos x}{x^{1/3}}\bigg|_{x=1}^{x=\infty}-\dfrac{1}{3}\int_{1}^{\infty}\dfrac{\cos x}{x^{4/3}}dx\\ &=\cos 1-\dfrac{1}{3}\int_{1}^{\infty}\dfrac{\cos x}{x^{4/3}}dx, \end{align*} and $\left|\dfrac{\cos x}{x^{4/3}}\right|\leq\dfrac{1}{x^{4/3}}$ on $[1,\infty)$ and hence $\displaystyle\int_{1}^{\infty}\dfrac{\cos x}{x^{4/3}}dx$ converges absolutely.