Improper integral $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$

Question

If I have to solve $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ it is right to write: $$-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}$$ and so the integral is divergent? I have use the fact that $-\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \frac{-1}{x^{\frac{1}{3}}}$.

If I have instead $\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$, so without minus sign, what can I do to prove the divergence? Is it true that from the following I can deduce the divergence? $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx=-\int_{3}^{+\infty}\frac{-\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq -\int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}\, dx $$ My doubt arises since I have seen here that a similar integral here (Solution verification of the improper integral $\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt$) is solved in a more complicate way...

Answer 1

$$\int_{2n\pi}^{2n\pi+\pi}\frac{\sin^4x}{x^{1/3}}dx\geq\frac{1}{\sqrt[3]{2n\pi+\pi}}\int_{2n\pi}^{2n\pi+\pi}\sin^4xdx\\ =\frac{1}{\sqrt[3]{2n\pi+\pi}}\int_{0}^{\pi}\sin^4xdx\\ =\frac{2}{\sqrt[3]{2n\pi+\pi}}\int_{0}^{\frac\pi2}\sin^4xdx =\frac{1}{\sqrt[3]{2n\pi+\pi}}\cdot\frac{3\pi}{8}.$$ So $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq\frac{3\pi}{8}\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{2n\pi+\pi}}=+\infty.$$

Answer 2

Hint:

Let us avoid the confusion raised by the minus sign, using the fact that if $-I$ diverges, so does $I$.

Now

$$\frac{\sin^4x}{x^{1/3}}<\frac1{x^{1/3}}$$ is of no help because it only gives an upper bound on the function and the function could remain much smaller. What you need is a lower bound.

A correct way to solve is to write for instance

$$\frac{\sin^4x}{x^{1/3}}>\frac1{\color{red}4\,x^{1/3}}$$ which holds in a constant fraction of every period, and this will be enough to justify divergence.